Trigonometry — Class 10 Maths PYQs

(b) $\frac{1}{5}$

(C) $3$

(a) $\frac{17}{7}$

(B) $\frac{y}{\sqrt{x^2 + y^2}}$

(C) $\frac{\sqrt{y^2 - x^2}}{x}$

(A) $\frac{2}{5}$

(B) $\frac{\sqrt{10}}{3}$

(a) $\frac{\sqrt{5}}{2}$

(D) $\frac{4}{3}$

(B) $\frac{12}{13}$

(C) $3$

(C) $\frac{q}{\sqrt{p^2 + q^2}}$

(D) $0$

(D) $\frac{4}{\sqrt{15}}$

(c) $\frac{\sqrt{b^2-a^2}}{b}$

(d) $\frac{1}{2\sqrt{2}}$

(B) $\frac{3}{4}$ (1 Mark)

(C) $\frac{3}{\sqrt{10}}$

(B) $2\sqrt{2}$

(D) $\frac{82}{80}$ (1 Mark)

(C) $\frac{65}{63}$

(c) $\sqrt{2}$

(b) $\frac{\sqrt{5}}{3}$

(c) $\frac{1}{\sqrt{5}}$

(C) $2\sqrt{2}$

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(C) 0

(C) $\frac{\sqrt{3}}{2}$

(A) $\frac{1}{4\sqrt{5}}$

(A) $\frac{1}{4\sqrt{3}}$

(A) $\frac{1}{4\sqrt{3}}$

Answer (A) $\frac{1}{\sqrt{6}}$

Answer (C) $\sqrt{10}$

(D) $\frac{4}{3}$

$$\begin{aligned}& \sec^2 \theta = 1 + \frac{1}{7} = \frac{8}{7} \\ & \cot \theta = \sqrt{7} \Rightarrow \text{cosec}^2 \theta = 1 + 7 = 8 \\ & \therefore \text{LHS} = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} \\ & = \frac{3}{4} = \text{RHS}\end{aligned}$$

$\text{sin } A = \frac{12}{13}$, $\text{cos } A = \frac{5}{13}$
Hence $\frac{2 \text{sin } A - 3 \text{cos } A}{4 \text{sin } A - 9 \text{cos } A} = \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13}}{4 \times \frac{12}{13} - 9 \times \frac{5}{13}} = 3$

$$\begin{aligned}& \sin A = \frac{3}{5} \Rightarrow \tan A = \frac{3}{4} \\ & \cos B = \frac{12}{13} \Rightarrow \tan B = \frac{5}{12} \\ & \tan A + \tan B = \frac{3}{4} + \frac{5}{12} = \frac{14}{12}\end{aligned}$$ or $\frac{7}{6}$

$\sin \alpha = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\sin \alpha = \frac{6}{3+9} = \frac{6}{12} = \frac{1}{2}$

$\cos A = \sqrt{1-\sin^2 A} = \sqrt{1-y^2}$
$\tan A = \frac{\sin A}{\cos A} = \frac{y}{\sqrt{1-y^2}}$

$\tan \theta = \frac{24}{7} = \frac{P}{B}$
Getting $\sin \theta = \frac{24}{25}$ and $\cos \theta = \frac{7}{25}$ (I) (1½ Mark)
$\therefore \sin \theta + \cos \theta = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}$ (II) (1/2 Mark)

$(\tan \theta + \frac{1}{\tan \theta})^2 = (2)^2$ (1/2 Mark)
$\tan^2 \theta + \frac{1}{\tan^2 \theta} + 2 = 4$ (1 Mark)
$\Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} = 2$ (1/2 Mark)

$\sin A = \frac{4}{5}$ (1 Mark)
$\cos A = \frac{3}{5}$ (1 Mark)

$BC = \sqrt{(-8 - 1)^2 + (7 - 1)^2} = \sqrt{117} = 3\sqrt{13}$ (I Mark)
$AB = \sqrt{(3 - 1)^2 + (4 - 1)^2} = \sqrt{13}$ (II Mark)
$\tan A = \frac{BC}{AB} = \frac{3\sqrt{13}}{\sqrt{13}} = 3$ (III Mark)

$4 \tan A = 3$
$\tan A = \frac{3}{4}$ ($\frac{1}{2}$ Mark)
$\Rightarrow \cot A = \frac{4}{3}$ ($\frac{1}{2}$ Mark)
$\therefore \text{cosec}^2 A = 1 + (\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$ (1 Mark)
$\frac{\text{cosec}^2 A + 1}{\text{cosec}^2 A - 1} = \frac{\frac{25}{9} + 1}{\frac{25}{9} - 1} = \frac{\frac{34}{9}}{\frac{16}{9}} = \frac{34}{16}$ or $\frac{17}{8}$ (1 Mark)

(a) $\text{cosec } A = \frac{25}{24}$ and $\tan A = \frac{24}{7}$

$LHS = \sin (60^\circ + 30^\circ) = \sin 90^\circ = 1$
$RHS = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = 1 = LHS$

$\tan \theta = \frac{1}{\sqrt{7}}$
$\Rightarrow \sec^2\theta = \frac{8}{7}$ and $\text{cosec}^2\theta = 8$
$\therefore \text{LHS} = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} = \frac{3}{4} = \text{RHS}$

(i) Hypotenuse $AC = 17$ cm $\implies \sin^2 A - \cos^2 A = (\frac{15}{17})^2 - (\frac{8}{17})^2 = \frac{161}{289}$
(ii) $\frac{1}{\cos^2 A} - \frac{1}{\cot^2 A} = \frac{17^2}{8^2} - \frac{15^2}{8^2} = 1$
(iii) (a) $\frac{2 \tan A}{1 + \tan^2 A} = \frac{2 \times \frac{15}{8}}{1 + \frac{15^2}{8^2}} = \frac{240}{289}$
$2 \sin A \cos A = 2 \times \frac{15}{17} \times \frac{8}{17} = \frac{240}{289}$ Hence they are equal
Note : Marks should be awarded to the alternate solution as well : $\frac{2 \tan A}{\sec^2 A} = \frac{2 \times \frac{\sin A}{\cos A}}{\frac{1}{\cos^2 A}} = 2 \sin A \cos A$
OR
(iii) (b) $\frac{\tan^2 A - \sec^2 A}{\cot^2 A - \csc^2 A} = \frac{(\frac{15}{8})^2 - (\frac{17}{8})^2}{(\frac{8}{15})^2 - (\frac{17}{15})^2} = 1$

(D) $\tan^2\theta- \sec^2 \theta = 1$

(A) $\operatorname{cosec} \theta \sec \theta$

(A) $\text{cosec } \theta \text{ sec } \theta$

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(B) $\sqrt{2} - 1$

(C) Assertion (A) is true, but Reason (R) is false

(C) Assertion (A) is true, but Reason (R) is false

Answer (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(b) LHS = $\tan (60^{\circ} - 30^{\circ}) = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ (1/2 Mark)
RHS = $\frac{\tan 60^{\circ} - \tan 30^{\circ}}{1 + \tan 60^{\circ} \tan 30^{\circ}} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} = \frac{\frac{3-1}{\sqrt{3}}}{1+1} = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}}$ (1/2 Mark)
Hence verified

(A) $-1$

(C) 0

(a) $\sin 60^{\circ}$

(b) $\cos 60^\circ$

(b) $\frac{\sqrt{3}}{2}$

$\frac{\sec^2 45^\circ - \tan^2 45^\circ}{\sin^2 45^\circ} = \frac{(\sqrt{2})^2-(1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{2-1}{\frac{1}{2}} = 2$

(A) $4$

(B) $\cos 60^\circ$

(B) $-\tan 60^\circ$

(B) $-\sin 30^{\circ}$

(C) 2:1

(C) $x < y$

(C) $3$

(b) $\frac{5}{4}$

(B) $\frac{4}{\sqrt{3}}$ (1 Mark)

(C) $1$

(C) $\frac{1}{4}$

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(A) $-1$

(C) $1$

(D) $-\sqrt{3}$

(B) $\frac{5}{3}$

(A) -1

(A) $\sin 45^{\circ}$

(A) $\sec 30^\circ$

(B) $\cos 60^\circ$

(D) $\sec x$

$\sin 15^\circ = \sin (45^\circ - 30^\circ)$ (1/2 Mark)
$= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$ (1/2 Mark)
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}$ (1 Mark)
$= \frac{\sqrt{3}-1}{2\sqrt{2}}$ or $\frac{\sqrt{6}-\sqrt{2}}{4}$ (1/2 Mark)

(B) $\cos 0^\circ$

$\cos 90^\circ$

(C) $-\frac{3}{4}$

(A) $1$

(B) $\frac{3}{2}$

$\frac{5\sin^2 45^{\circ}-3\tan^2 30^{\circ}}{2\sec^2 30^{\circ}}$
$5\times(\frac{1}{\sqrt{2}})^2 - 3\times(\frac{1}{\sqrt{3}})^2$ (1/2 Mark)
$\frac{5}{2} - \frac{3}{3}$
$2 \times (\frac{2}{\sqrt{3}})^2$
$\frac{9}{16}$ (1/2 Mark)

(A) $3$

$4 \cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + p = \frac{3}{4}$
$\Rightarrow 4(1)^2 - (2)^2 + (\frac{\sqrt{3}}{2})^2 + p = \frac{3}{4}$
$\Rightarrow 4 - 4 + \frac{3}{4} + p = \frac{3}{4}$
$\Rightarrow p = 0$

$\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} = \frac{5(1/2)^2 + 4(2/\sqrt{3})^2 - (1)^2}{1}$
$= \frac{5/4 + 16/3-1}{1} = \frac{67}{12}$

$$\begin{aligned}& \frac{5}{\cot^2 30^\circ} + \frac{1}{\sin^2 60^\circ} - \cot^2 45^\circ + 2 \sin^2 90^\circ \\ & = \frac{5}{(\sqrt{3})^2} + \frac{1}{(\sqrt{3}/2)^2} - (1)^2 + 2(1)^2 = \frac{5}{3} + \frac{4}{3} - 1 + 2 \\ & = \frac{9}{3} + 1 = 4 \\ & = 3 + 1 = 4\ \text{OR}\end{aligned}$$

$$\begin{aligned}& 2 \sec^2 45^\circ + 3 \text{cosec}^2 45^\circ - 2 \sin 45^\circ \cos 45^\circ \\ & = 2(\sqrt{2})^2 + 3(\sqrt{2})^2 - 2(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) \\ & = 4 + 6 - 1 = 9\end{aligned}$$

$3 \tan^2 60^\circ - x \sin^2 45^\circ + \frac{3}{4} \sec^2 30^\circ = 2 cosec^2 30^\circ$
$\Rightarrow 3(\sqrt{3})^2 - x(\frac{1}{\sqrt{2}})^2 + \frac{3}{4}(\frac{2}{\sqrt{3}})^2 = 2(2)^2$
$\Rightarrow 9 - \frac{x}{2} + 1 = 8$
$\Rightarrow x = 4$

$2\sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2} + 2\sqrt{3} \times \frac{\sqrt{3}}{2}$
$= 4$

LHS = $\sin (60^\circ + 30^\circ) = \sin 90^\circ = 1$
RHS = $\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ$
$= \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = 1$
$\therefore$ LHS = RHS

$2\sin^2 30^\circ \sec 60^\circ + \tan^2 60^\circ$
$= 2 \times (\frac{1}{2})^2 \times 2 + (\sqrt{3})^2$
$= 2 \times \frac{1}{4} \times 2 + 3$
$= 1 + 3$
$= 4$

$\frac{\cos 45^\circ + \sin 60^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$
$= \frac{\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}}{\frac{2}{\sqrt{3}} + 2}$
$= \frac{2\sqrt{3}+3\sqrt{2}}{4\sqrt{2}(1+\sqrt{3})}$

$\frac{5\tan60^{\circ}}{(\sin^2 60^{\circ} + \cos^2 60^{\circ})\tan30^{\circ}}$
$= \frac{5 \times \sqrt{3}}{1 \times \frac{1}{\sqrt{3}}}$
$=15$

$$\begin{aligned}& 5(\frac{1}{2})^2+4(\frac{2}{\sqrt{3}})^2-(1)^2 \\ & (\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2 \\ & = \frac{67}{12}\end{aligned}$$

$$\begin{aligned}& 2\times\frac{1}{\sqrt{3}}\times 2\times 1 \\ & \frac{1 - \frac{3}{4}}{} \\ & = \frac{16}{\sqrt{3}} \text{ or } \frac{16\sqrt{3}}{3}\end{aligned}$$

$x(\frac{1}{2}) + y(1) + \frac{1}{2} - 1 = 5 \implies x + 2y = 11$

$\frac{(\sqrt{3})^2}{(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 2$

$4k = (\sqrt{3})^2 - 2(2)^2 - 2(\frac{1}{\sqrt{3}})^2$
$= 3 - 2(4) - 2(\frac{1}{3})$
$= 3 - 8 - \frac{2}{3}$
$= -5 - \frac{2}{3}$
$= \frac{-15-2}{3} = \frac{-17}{3}$
$k = \frac{-17}{12}$

$\frac{3 \sin 30^\circ-4 \sin^3 30^\circ}{2 \sin^2 50^\circ +2 \cos^2 50^\circ}$
$= \frac{3\times\frac{1}{2}-4\times(\frac{1}{2})^3}{2 (\sin^2 50^\circ+\cos^2 50^\circ)}$
$= \frac{\frac{3}{2}-\frac{1}{2}}{2\times 1}$
$= \frac{1}{2}$

$\sin 15^\circ = \sin(45^\circ - 30^\circ)$
$= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}$
$= \frac{\sqrt{3}-1}{2\sqrt{2}}$ or $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\frac{5(\frac{1}{\sqrt{3}})^2 + 3(\frac{1}{\sqrt{2}})^2 - 4(\frac{1}{2})^2}{\sqrt{3} .(\frac{\sqrt{3}}{2}). \frac{1}{2} + (1)^2} = \frac{26}{21}$

$3\times(\frac{\sqrt{3}}{2})^2 - 6\times(2)^2$ (1/2 Mark)
$\frac{3 \times \frac{3}{4} - 6 \times 4}{(\sqrt{3})^2}$ (1/2 Mark)
= $\frac{\frac{9}{4} - 24}{3}$ (1/2 Mark)
= $\frac{\frac{9-96}{4}}{3} = \frac{-87}{12}$ or $-\frac{29}{4}$ (1/2 Mark)

$= \frac{1-2\times(\frac{1}{\sqrt{3}})^2-(\sqrt{2})^2}{(\frac{\sqrt{3}}{2})^2}$ (I) (1½ Mark)
$= -\frac{20}{9}$ (II) (1/2 Mark)

$\frac{(\frac{\sqrt{3}}{2})^3 - \frac{1}{\sqrt{3}}}{(\frac{1}{\sqrt{2}})^2}$ (1/2 Mark)
= $\frac{\frac{3\sqrt{3}}{8} - \frac{1}{\sqrt{3}}}{\frac{1}{2}}$ (1/2 Mark)
= $\frac{\frac{9-8}{8\sqrt{3}}}{\frac{1}{2}}$ (1/2 Mark)
= $\frac{1}{8\sqrt{3}} \times 2 = \frac{1}{4\sqrt{3}}$ or $\frac{\sqrt{3}}{12}$ (1/2 Mark)

$\frac{2 \cos 30^\circ - \cot^3 60^\circ}{\tan 30^\circ} = \frac{2 \times (\frac{\sqrt{3}}{2}) - (\frac{1}{\sqrt{3}})^3}{\frac{1}{\sqrt{3}}}$ (I Mark)
$= \frac{8}{3}$ (II Mark)

$\frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$ [1 1/2 marks]
$= \frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$ or $\frac{3\sqrt{2} - \sqrt{6}}{8}$ [1/2 mark]

$\frac{\sin^2 45^{\circ}}{\text{cosec}^2 30^{\circ} - \tan^2 45^{\circ}} = \frac{(\frac{1}{\sqrt{2}})^2}{(2)^2 - (1)^2}$ [1 1/2 marks]
$= \frac{1}{6}$ [1/2 mark]

(a) $2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - (1)^2 = \frac{7}{4}$

$LHS = \cos 60^\circ = \frac{1}{2}$
$RHS = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{1}{2} = LHS$

$\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2}}$ ($1\frac{1}{2}$ marks)
$= \frac{2\sqrt{3}}{5\sqrt{2}} \text{ or } \frac{\sqrt{6}}{5}$ ($\frac{1}{2}$ mark)

$LHS = \sin 60^\circ = \frac{\sqrt{3}}{2}$ ($\frac{1}{2}$ mark)
$RHS = \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\sqrt{3}}{2} = LHS$ ($1 + \frac{1}{2}$ marks)

(A) Given expression = $(\frac{1}{2})^2 - (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{3}})^2$ (1 Mark)
$= \frac{1}{12}$ (1 Mark)

$\frac{2 \sin^2 60^\circ + \cos^2 60^\circ}{\tan^2 30^\circ}$
$= \frac{2 \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2}{\left( \frac{1}{\sqrt{3}} \right)^2}$ (3$\times$1/2 Mark)
$= \frac{21}{4}$ (1/2 Mark)

$\Delta PQR$ is an isosceles right triangle
$:: \angle P= \angle R = 45^{\circ}$ (1/2 Mark)
$\sec^2 P + \text{cosec}^2 R = \sec^2 45^{\circ} + \text{cosec}^2 45^{\circ}$
$= (\sqrt{2})^2 + (\sqrt{2})^2$
$= 4$ (1/2 Mark)

$\sin 45^{\circ} \cdot \cos 45^{\circ} + \text{cosec}^2 30^{\circ}$
$= \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + 2^2$ (3× 1/2 Mark)
$= \frac{1}{2} + 4 = \frac{9}{2}$ (1/2 Mark)

$\sec 60^\circ \cdot \cos^2 30^\circ + \sin^2 45^\circ$
$= 2 \times (\frac{\sqrt{3}}{2})^2 + (\frac{1}{\sqrt{2}})^2$ (1/2+1/2+1/2 Mark)
$= 2 \times \frac{3}{4} + \frac{1}{2}$ (1/2 Mark)
$= \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$

$\sec A = \sqrt{2} \Rightarrow A = 45^{\circ}$, $\tan B = \sqrt{3} \Rightarrow B = 60^{\circ}$ (1/2 Mark + 1/2 Mark)
$2 \sin A \cos B = 2 \sin 45^{\circ} \cos 60^{\circ} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$ (1 Mark)

$\frac{4\cos^3 60^\circ + \text{cosec } 30^\circ}{\tan^2 30^\circ} = \frac{4 \times (\frac{1}{2})^3 + 2}{(\frac{1}{\sqrt{3}})^2}$ (1 Mark)
$= \frac{4 \times \frac{1}{8} + 2}{\frac{1}{3}} = \frac{\frac{1}{2} + 2}{\frac{1}{3}} = \frac{\frac{5}{2}}{\frac{1}{3}} = \frac{15}{2}$ (½ Mark)

$\frac{(\sqrt{3})^2+4(\frac{1}{\sqrt{2}})^2+3(2)^2+5(0)^2}{2+2-(\sqrt{3})^2}$ (3 Marks)
$= \frac{3+2+12+0}{4-3}$ (1 Mark)
$= 17$ (1 Mark)

(d) $30^\circ$

(B) $45^\circ$

(a) $\frac{1}{\sqrt{2}}$

(C) $1$

(A) $\sqrt{3}$

(C) $1$

(A) $\frac{1}{2\sqrt{2}}$

(A) $30^\circ$

(D) $60^\circ$

(B) $30^\circ$

(B) $\frac{1}{2}$

(A) $\frac{1}{\sqrt{2}}$

(D) $10^{\circ}$

(D) $5^\circ$

(D) $10^{\circ}$

(d) $30^\circ$

(c) $45^{\circ}$

(D) $\frac{1}{4}$

(D) $0^{\circ}$

(C) $45^\circ$

(D) $30^\circ$

(d) $30^\circ$

(d) $(\frac{22}{2})^\circ$

(b) $45^\circ$

(A) $0^{\circ}$

$A = 45^\circ, B = 60^\circ$
$\cos A \cos B + \sin A \sin B$
$= \cos 45^\circ \cos 60^\circ + \sin 45^\circ \sin 60^\circ$
$= \frac{1}{\sqrt{2}} \times \frac{1}{2} + \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}$
$= \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1+\sqrt{3}}{2\sqrt{2}}$

$\text{cosec} \alpha = \frac{1}{\sin \alpha} = \sqrt{2}$
$\text{cosec} \beta = \sqrt{1 + \cot^2 \beta} = \sqrt{1+3} = 2$
$\therefore \text{cosec} \alpha + \text{cosec} \beta = \sqrt{2} + 2$ or $\sqrt{2} (\sqrt{2} + 1)$

$\sin (A - B) = 0 \Rightarrow A-B=0^\circ$
$\cos (A + B) = \frac{1}{2} \Rightarrow A + B = 60^\circ$
$\Rightarrow A = 30^\circ, B = 30^\circ$

$$\begin{aligned}& \sin \theta = \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = 1 \Rightarrow \tan \theta = 1 \Rightarrow \cot \theta = 1 \\ & \tan^2 \theta + \cot^2 \theta – 2 = (1)^2 + (1)^2 – 2 = 0\end{aligned}$$

$\cos(A + B) = \frac{1}{2} \Rightarrow A + B = 60^\circ$ ... (i)
$\tan(A - B) = \frac{1}{\sqrt{3}} \Rightarrow A - B = 30^\circ$ ... (ii)
Solving (i) and (ii), we get $A = 45^\circ$ and $B = 15^\circ$
$\Rightarrow \sec(2A - 3B) = \sec(90^\circ - 45^\circ)$
$= \sec 45^\circ = \sqrt{2}$

$\sin(A + B) = \frac{\sqrt{3}}{2} \Rightarrow A + B = 60^\circ \dots (1)$
$\cos(A - B) = 1 \Rightarrow A - B = 0^\circ \dots (2)$
Solving $(1)$ and $(2)$, we get $A = B = 30^\circ$

$$\begin{aligned}& \sin (A - B) = \sin 30^\circ \\ & A - B = 30^\circ --------(i) \\ & \cos (A+B) = \cos 60^\circ \\ & A + B = 60^\circ ---------(ii) \\ & \text{Solving (i) and (ii)} \\ & A = 45^\circ, B = 15^\circ\end{aligned}$$

$\sin (2A + 3B) = 1 \Rightarrow 2A + 3B = 90^{\circ}$ --- (1)
$\cos (2A - 3B) = \frac{\sqrt{3}}{2} \Rightarrow 2A - 3B = 30^{\circ}$ --- (2)
Solving (1) and (2), we get
A = $30^{\circ}$ and B = $10^{\circ}$

$\tan A = \sqrt{3} = \tan 60^\circ$
$\Rightarrow A = 60^\circ$
$\frac{\sin^2 A}{1+\cos^2 A} = \frac{\sin^2 60^\circ}{1+\cos^2 60^\circ}$
$= \frac{(\frac{\sqrt{3}}{2})^2}{1+(\frac{1}{2})^2}$
$= \frac{\frac{3}{4}}{1+\frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}}$
$= \frac{3}{5}$

$\tan(A + 2B) = \sqrt{3} \Rightarrow A + 2B = 60^\circ$ (I) (1/2)
$ \sin(2A + B) = \frac{1}{\sqrt{2}} \Rightarrow 2A + B = 45^\circ$ (II) (1/2)
On solving above equations, $A = 10^\circ$, $B = 25^\circ$ (III) (1/2+1/2)

$\sec(2A - B) = \sqrt{2} \Rightarrow 2A - B = 45^\circ$ (I Mark)
$\operatorname{cosec}(A + B) = 2 \Rightarrow A + B = 30^\circ$ (II Mark)
On solving, $A = 25^\circ, B = 5^\circ$ (III Mark)

$\sin (A + 2B) = 1$ (1/2 Mark)
$A + 2B = 90^\circ$ (1/2 Mark)
Using $A = 3B$, we get
$B = 18^\circ$ (1/2 Mark)
$A = 54^\circ$ (1/2 Mark)

$\sin (A - B) = \frac{1}{2} \Rightarrow A-B = 30^{\circ}$ ..........(i) (1/2 Mark)
$\tan (A + B) = \sqrt{3} \Rightarrow A + B = 60^{\circ}$ ..........(ii) (1/2 Mark)
Solving (i) and (ii) to get $A = 45^{\circ}$, $B = 15^{\circ}$ (1 Mark)

Getting $A = 30^\circ$ and $B = 45^\circ$ (1½)
$A + B = 75^\circ$ (½)

$\sin(A + 2B) = \frac{\sqrt{3}}{2} \Rightarrow A + 2B = 60^\circ$ (1/2 Mark)
$\cos(A + 4B) = 0 \Rightarrow A + 4B = 90^\circ$ (1/2 Mark)
Solving to get $A = 30^\circ$, $B = 15^\circ$ (1 Mark)

$\sin(2A + B) = \frac{\sqrt{3}}{2} \implies 2A + B = 60^\circ$ --- (1)
$\tan(A + 2B) = 1 \implies A + 2B = 45^\circ$ --- (2)
Solving (1) $\&$ (2), we get $A = 25^\circ$ and $B = 10^\circ$
$\cot(4A - 7B) = \cot 30^\circ = \sqrt{3}$

(a) $A + B = 45^{\circ}$
$A - B = 30^{\circ}$
Solving and getting $A = 37.5^{\circ}$ and $B = 7.5^{\circ}$

$\sin (A + 2B) = 1 \Rightarrow A + 2B = 90^\circ$ --------(i) (1/2 Mark)
$\cos (2A + B) = \frac{1}{2} \Rightarrow 2A + B = 60^\circ$ --------(ii) (1/2 Mark)
Solving (i) and (ii) we get $A = 10^\circ$ and $B = 40^\circ$ (1 Mark)
$\tan (B-A) = \tan (40^\circ – 10^\circ) = \tan 30^\circ = \frac{1}{\sqrt{3}}$ (1 Mark)

(d) $- 1$

(c) $2$

(b) $-1$

(c) $\frac{\sqrt{1 + \cot^2 \theta}}{\cot \theta}$

(c) $\sec^2\theta – \tan^2\theta = 1$

(b) $1$

(d) $3$

(D) -1

(A)

(D) $2 \cos^2 A - 1$

(C) $1$

(c) $\frac{1}{m}$

(A) Both Assertion (A) and (R) are true. Reason (R) is the correct explanation of Assertion (A)

(A) $36$

(D) $\cos \theta$

(A) $1$

(A) $1$

(C) $-1$

(B) $1$

(A) $\cot \theta$

(c) $\sec^2 \theta = 1 + \tan^2 \theta$

(C) Assertion (A) is true, but Reason (R) is false.

(A) $\tan^2 A$

(D) $\frac{\sqrt{13}}{2}$

(B) $\sec^2 A$

(C) $\operatorname{cosec} A$

(D) $- 1$

(B) $1$

(C) $3$

(B) $2$

(A) $0$

(B) $-1$

(D) Assertion (A) is false, but Reason (R) is True.

(a) $\frac{1}{2}$

$\sin \theta + \cos \theta = \sqrt{3}$
squaring both sides
$\sin^2 \theta + \cos^2 \theta + 2 \sin\theta \cos \theta = 3$
$\Rightarrow 1 + 2 \sin \theta \cos \theta = 3$
$\Rightarrow \sin \theta \cos \theta = 1$

$$\begin{aligned}& \sin \theta + \sin^2 \theta = 1 \\ & \Rightarrow \sin \theta = 1 - \sin^2 \theta = \cos^2 \theta \\ & \therefore \cos^2 \theta + \cos^4 \theta = \cos^2 \theta (1 + \cos^2 \theta) \\ & = \sin \theta (1 + \sin \theta) \\ & = \sin \theta + \sin^2 \theta= 1\end{aligned}$$

$\cos A + \cos^2 A = 1 \Rightarrow \cos A = 1 - \cos^2 A = \sin^2 A$
$\therefore \sin^2 A + \sin^4 A = \cos A + \cos^2 A (\because \sin^2 A = \cos A)$
$= 1$

$$\begin{aligned}& \sin\theta - \cos\theta = 0 \Rightarrow \sin\theta = \cos\theta \Rightarrow \tan\theta = 1 \\ & Rightarrow \theta = 45^\circ \\ & sin^4 45^\circ + \cos^4 45^\circ = (\frac{1}{\sqrt{2}})^4 + (\frac{1}{\sqrt{2}})^4 \\ & = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\end{aligned}$$

$(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = x + \tan^2 A + \cot^2 A$
$\Rightarrow \sin^2 A + \text{cosec}^2 A + 2 + \cos^2 A + \sec^2 A + 2 = x + \tan^2 A + \cot^2 A$
$\Rightarrow 1 + 2 + 2 + 1 + \cot^2 A + 1 + \tan^2 A = x + \tan^2 A + \cot^2 A$
$\therefore x = 7$

$(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = x + \tan^2 A + \cot^2 A$
$\implies \sin^2 A + \csc^2 A + 2 + \cos^2 A + \sec^2 A + 2 = x + \tan^2 A + \cot^2 A$
$\implies 1 + 2 + 2 + 1 + \cot^2 A + 1 + \tan^2 A = x + \tan^2 A + \cot^2 A$
$\therefore x = 7$

$\sin^2 A + \cos^2 A = 1$.
Dividing both sides by $\cos^2 A$,
we get $\frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\cos^2 A} = \frac{1}{\cos^2 A}$ ($\frac{1}{2}$ mark).
$\tan^2 A + 1 = \sec^2 A$ ($\frac{1}{2}$ mark).
$\tan^2 A + 1 = (\frac{5}{3})^2$ ($\frac{1}{2}$ mark).
$\tan A = \frac{4}{3}$ ($\frac{1}{2}$ mark).

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta}$ (I) (1 Mark)
$= \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta$ (II) (1/2 Mark)
$= (\frac{7}{8})^2 = \frac{49}{64}$ (III) (1/2 Mark)

$\cos A = \sqrt{1 - \sin^2 A}$ (1 Mark)
$\tan A = \frac{\sin A}{\cos A} = \frac{\sin A}{\sqrt{1-\sin^2 A}}$ (1 Mark)

$7 \sin^2 A + 3 \cos^2 A = 4$ ($\frac{1}{2}$ Mark)
$\Rightarrow 7 \sin^2 A + 3(1 - \sin^2 A) = 4$ ($\frac{1}{2}$ Mark)
$\Rightarrow 4 \sin^2 A = 1$ ($\frac{1}{2}$ Mark)
$\Rightarrow \sin A = \frac{1}{2}$ ($\frac{1}{2}$ Mark)
$\Rightarrow A = 30^{\circ}$ ($\frac{1}{2}$ Mark)
$\therefore \tan A = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ ($\frac{1}{2}$ Mark)

$3A = 90^\circ \Rightarrow A = 30^\circ$.
$$\begin{aligned}& \cos 2A - \tan^2 45^\circ \\ & = \cos 60^\circ - \tan^2 45^\circ \\ & = \frac{1}{2} - 1 \\ & = -\frac{1}{2}\end{aligned}$$. ($\frac{1}{2} + 1\frac{1}{2}$ marks)

$$\begin{aligned}& \left( \frac{1}{\cos A} + \frac{\sin A}{\cos A} \right) (1 - \sin A) \\ & = k \cos A \Rightarrow 1 - \sin^2 A = k \cos^2 A \Rightarrow \cos^2 A \\ & = k \cos^2 A \Rightarrow k = 1\end{aligned}$$. ($\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$ marks)

(a) $\triangle ABC$ is an isosceles right triangle
$\therefore \angle A = \angle C = 45^\circ$ (1/2 Mark)
$\sin^2 A + \cos^2 C = \sin^2 45^\circ + \cos^2 45^\circ$
$= \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2$ (1/2+1/2 Mark)
$= 1$ (1/2 Mark)

$\sin A + \sin^2 A = 1$
$\implies \sin A = \cos^2 A$ (1/2 Mark)
$\sin^2 A = \cos^4 A$ (On squaring both sides) (1/2 Mark)
$1 - \cos^2 A = \cos^4 A \implies \cos^2 A + \cos^4 A = 1$ (1 Mark)
LHS $= \tan^2 A \cdot \sec^2 A = \frac{\sin^2 A}{\cos^2 A} \times \frac{1}{\cos^2 A} = \frac{\sin^2 A}{\sin^2 A} \times 1 = 1 = \text{RHS}$ ($:: \cos^4 A = \sin^2 A$) (1 Mark)

(A) $1$

(A) $a^2b^2$

$m^2 + n^2 = (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2$
$= a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2 \theta + \cos^2 \theta)$
$= a^2 + b^2$

LHS $= \frac{\sqrt{\sec A-1}}{\sqrt{\sec A+1}} + \frac{\sqrt{\sec A+1}}{\sqrt{\sec A-1}}$
$= \frac{(\sec A-1) + (\sec A+1)}{\sqrt{(\sec A+1)(\sec A-1)}}$
$= \frac{2 \sec A}{\sqrt{\sec^2 A-1}}$
$= \frac{2 \sec A}{\sqrt{\tan^2 A}}$
$= \frac{2 \sec A}{\tan A}$
$= \frac{2/\cos A}{\sin A/\cos A}$
$= \frac{2}{\sin A}$
$= 2 \operatorname{cosec} A = \text{RHS}$

$$\begin{aligned}& \tan \theta + \sec \theta = m \dots (i) \\ & Therefore, \sec \theta - \tan \theta = \frac{1}{m} \dots (ii) \\ & Adding (i)\end{aligned}$$ and $(ii)$ to get
$$\begin{aligned}& 2 \sec \theta = m + \frac{1}{m} \\ & \sec \theta = \frac{m^2+1}{2m}\end{aligned}$$

$$\begin{aligned}& (\tan A + \cot A)^2 = 36 \\ & \tan^2 A + \cot^2 A + 2\tan A \cot A = 36 \\ & \tan^2 A + \cot^2 A = 34 \\ & \therefore \tan^2 A + \cot^2 A - 4 = 30\end{aligned}$$

$m^2 = a^2 \sec^2 \theta + b^2 \tan^2 \theta + 2ab \sec \theta \tan \theta$ ($\frac{1}{2}$ mark).
$n^2 = b^2 \sec^2 \theta + a^2 \tan^2 \theta + 2ab \sec \theta \tan \theta$ ($\frac{1}{2}$ mark).
$m^2 - n^2 = a^2(\sec^2 \theta - \tan^2 \theta) + b^2(\tan^2 \theta - \sec^2 \theta)$ ($\frac{1}{2}$ mark).
$\Rightarrow m^2 - n^2 = a^2 - b^2$ or $a^2 + n^2 = m^2 + b^2$ ($\frac{1}{2}$ mark).

(a) LHS = $$\begin{aligned}& (\text{cosec} \theta + \sin \theta) (\text{cosec} \theta - \sin \theta) \\ & = \frac{(1+\sin^2 \theta)(1-\sin^2 \theta)}{\sin^2 \theta} \\ & = (1 + \sin^2 \theta) (\frac{\cos^2 \theta}{\sin^2 \theta}) \\ & = (\cot^2 \theta + \cos^2 \theta)\end{aligned}$$

LHS $= \sqrt{\frac{(1 - \sin \theta)}{(1 + \sin \theta)} \times \frac{(1 - \sin \theta)}{(1- \sin \theta)}}$ (1/2 Mark)
$= \sqrt{\frac{(1 - \sin \theta)^2}{(1- \sin^2 \theta)}}$ (1/2 Mark)
$= \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}}$ (1/2 Mark)
$= \frac{(1 - \sin \theta)}{\cos \theta} = \sec\theta - \tan\theta = RHS$ (1/2 Mark)

L.H.S. $= \frac{\tan \theta}{\sec^2 \theta} + \frac{\cot \theta}{cosec^2 \theta}$ (I) (1 Mark)
$= \frac{\sin \theta}{\cos \theta} \times \cos^2 \theta + \frac{\cos \theta}{\sin \theta} \times \sin^2 \theta$ (II) (1/2 Mark)
$= 2 \sin \theta \cos \theta = \text{R.H.S.}$ (III) (1/2 Mark)

$\sin A = \frac{1}{2} \Rightarrow A = 30^\circ$, $\tan B = \sqrt{3} \Rightarrow B = 60^\circ$ (1 Mark)
LHS = $\cos (30^\circ + 60^\circ) = \cos 90^\circ = 0$ (1/2 Mark)
RHS = $\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ$
= $\frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2} = 0$ (1/2 Mark)
... LHS = RHS

LHS = $\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1/2 Mark)
$= \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = 2 \operatorname{cosec} \theta = \text{RHS}$ (1/2 Mark)

$$\begin{aligned}& LHS = \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}} \\ & = \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} \\ & = \frac{\sin^2 A}{\cos A (\sin A - \cos A)} - \frac{\cos^2 A}{\sin A (\sin A - \cos A)} \\ & = \frac{1}{(\sin A - \cos A)} \left[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} \right] \\ & = \frac{1}{(\sin A - \cos A)} \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{\sin A \cos A} \\ & = \frac{1 + \sin A \cos A}{\sin A \cos A} \\ & = \frac{1}{\sin A \cos A} + 1 \\ & = 1 + \sec A cosec A = RHS\end{aligned}$$

LHS = $\frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1}$
Dividing Numerator and Denominator by $\cos \theta$,
$\frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec^2\theta-\tan^2\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta)-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{1+\tan \theta - \sec \theta}$
$\frac{(\tan \theta + \sec\theta) (1-\sec\theta+\tan\theta)}{1+\tan \theta - \sec \theta}$
$= (\tan \theta + \sec \theta)$
Multiplying & dividing by $(\sec \theta – \tan \theta)$
$= (\tan \theta + \sec \theta) \times \frac{(\sec \theta - \tan \theta)}{(\sec \theta - \tan \theta)}$
$= \frac{(\sec^2\theta-\tan^2\theta)}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = \text{RHS}$

LHS $= (\frac{1}{\sin A} - \sin A) (\frac{1}{\cos A} - \cos A)$
$= (\frac{1 - \sin^2 A}{\sin A}) (\frac{1 - \cos^2 A}{\cos A})$
$= \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A}$
$= \sin A \cos A$
RHS $= \frac{1}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}}$
$= \frac{1}{\frac{\cos^2 A + \sin^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{1}$
$= \sin A \cos A = \text{LHS}$

LHS = $$\begin{aligned}& 2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[(\sin^2\theta)^3 + (\cos^2\theta)^3] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[(\sin^2\theta + \cos^2\theta)(\sin^4\theta - \sin^2\theta \cos^2\theta + \cos^4\theta)] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = 2[\sin^4\theta + \cos^4\theta - \sin^2\theta \cos^2\theta] - 3(\sin^4\theta + \cos^4\theta) + 1 \\ & = -[\sin^4\theta + \cos^4\theta + 2 \sin^2\theta \cos^2\theta] + 1 \\ & = -(\sin^2\theta + \cos^2\theta)^2 + 1 \\ & = -1 + 1 = 0\end{aligned}$$

$LHS = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta}$
$= \frac{\sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta}{(1 + \cos \theta) \sin \theta}$
$= \frac{1 + 1 + 2 \cos \theta}{(1 + \cos \theta) \sin \theta}$
$= \frac{2 + 2 \cos \theta}{(1 + \cos \theta) \sin \theta}$
$= \frac{2 (1 + \cos \theta)}{(1 + \cos \theta) \sin \theta}$
$= \frac{2}{\sin \theta} = 2 cosec \theta = RHS$

$LHS = \frac{\cos^2 \theta}{1-\tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}$
$= \frac{\cos^2 \theta}{1-\frac{\sin \theta}{\cos \theta}} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}$ ($\frac{1}{2}$)
$= \frac{\cos^3 \theta}{\cos \theta - \sin \theta} - \frac{\sin^3 \theta}{\cos \theta - \sin \theta}$ (1)
$= \frac{(\cos \theta - \sin \theta) (\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta)}{(\cos \theta - \sin \theta)}$ (1)
$= 1 + \cos \theta \sin \theta = RHS$ ($\frac{1}{2}$)

LHS = $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \frac{\sin A (1 – 2 \sin^2 A)}{\cos A (2 \cos^2 A – 1)}$
$= \frac{\sin A[1 - 2(1 – \cos^2 A)]}{\cos A [2 \cos^2 A- - 1]} = \frac{\sin A[1 – 2 + 2 \cos^2 A]}{\cos A[2 \cos^2 A - 1]}$
$= \frac{\sin A[2 \cos^2 A - 1]}{\cos A [2 \cos^2 A – 1]} = \tan A = RHS$

LHS = $\sec A (1 – \sin A) (\sec A + \tan A)$
$= \frac{1}{\cos A} (1 – \sin A) (\frac{1}{\cos A} + \frac{\sin A}{\cos A})$
$= \frac{1}{\cos A} (1 – \sin A) (\frac{1 + \sin A}{\cos A})$
$= \frac{1 - \sin^2 A}{\cos^2 A} = \frac{\cos^2 A}{\cos^2 A} = 1=RHS$

LHS $= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$
$= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
Using $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \operatorname{cosec} \theta \sec \theta + 1$
$= 1 + \sec \theta \operatorname{cosec} \theta = \text{RHS}$

$LHS = \frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A} = \frac{\frac{\sin A}{\cos A}}{1+\frac{1}{\cos A}} - \frac{\frac{\sin A}{\cos A}}{1-\frac{1}{\cos A}}$ (I) (1 Mark)
$= \frac{\sin A}{\cos A+1} - \frac{\sin A}{\cos A-1}$ (II) (1/2 Mark)
$= \sin A (\frac{-2}{\sin^2A})$ (III) (1 Mark)
$= \frac{2}{\sin A} = 2 cosec A = RHS$ (IV) (1/2 Mark)

$LHS = \frac{1 + \sec A}{\sec A} = \frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$
$= 1 + \cos A$
$= \frac{(1 - \cos A)(1 + \cos A)}{(1-\cos A)}$
$= \frac{1- \cos^2 A}{1-\cos A}$
$= \frac{\sin^2 A}{1-\cos A} = RHS$

$\sin \theta + \cos \theta = p$, $\sec \theta + \text{cosec } \theta = q$
LHS = $q(p^2 - 1)$
$= (\sec \theta + \text{cosec } \theta)[(\sin \theta + \cos \theta)^2 - 1]$
$= (\frac{1}{\cos \theta} + \frac{1}{\sin \theta})[\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1]$
$= (\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}) [1 + 2 \sin \theta \cos \theta - 1]$
$= (\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}) (2 \sin \theta \cos \theta)$
$= 2(\sin \theta + \cos \theta)$
$= 2p = RHS$

LHS $= (\sin \theta + \cos \theta) (\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta})$
$= (\sin \theta + \cos \theta)(\frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta})$
$= \frac{(\sin \theta + \cos \theta).(1)}{\cos \theta \sin \theta}$
$= \sec \theta + \text{cosec } \theta = \text{RHS}$

LHS $= \frac{1+\tan^2 A}{1+\frac{1}{\tan^2 A}} = \frac{\tan^2 A (1+\tan^2 A)}{1+\tan^2 A} = \tan^2 A$
RHS $= \frac{(1-\tan A)^2}{(1-\frac{1}{\tan A})^2} = \frac{(1-\tan A)^2}{(\frac{\tan A-1}{\tan A})^2} = \frac{(1-\tan A)^2 \tan^2 A}{(1-\tan A)^2} = \tan^2 A$
$\therefore$ LHS $=$ RHS

$LHS = \text{cosec}^2 A + \sin^2 A + 2 \text{cosec } A \sin A + \sec^2 A + \cos^2 A + 2 \cos A \sec A$
$= (\sin^2 A + \cos^2 A) + (1 + \tan^2 A) + (1 + \cot^2 A) + 4$
$= 7 + \tan^2 A + \cot^2 A = RHS$

$LHS = (\frac{\sin \alpha}{\cos \alpha} + \frac{1}{\cos \alpha})^2 + (\frac{\sin \alpha}{\cos \alpha} - \frac{1}{\cos \alpha})^2$
$= (\frac{\sin \alpha+1}{\cos \alpha})^2 + (\frac{\sin \alpha-1}{\cos \alpha})^2$
$= \frac{\sin^2\alpha+2 \sin \alpha+1+\sin^2\alpha-2 \sin \alpha+1}{1- \sin^2\alpha}$
$= \frac{2(1+\sin^2\alpha)}{1- \sin^2\alpha} = RHS$

L.H.S. = $\sqrt{(1 + \text{tan}^2A) + (1 + \text{cot}^2A)}$
$= \sqrt{\text{tan}^2A + \text{cot}^2A + 2}$
$= \sqrt{(\text{tanA} + \text{cotA})^2}$
$= (\text{tanA} + \text{cotA})$ = R.H.S.

$L.H.S = \frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}$
$= \frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}$
$= \frac{\tan\theta(1-2\sin^2\theta)}{[2(1-\sin^2\theta)-1]}$
$= \frac{\tan\theta(1-2\sin^2\theta)}{(1-2\sin^2\theta)}$
$= \tan\theta = R.H.S.$

LHS $= \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A - \cos A)(\sin A + \cos A)} = \frac{\sin^2 A + \cos^2 A + 2 \sin A \cos A + \sin^2 A + \cos^2 A - 2 \sin A \cos A}{\sin^2 A - \cos^2 A} = \frac{1 + 1}{\sin^2 A - (1 - \sin^2 A)} = \frac{2}{2 \sin^2 A - 1} = \text{RHS}$.

L.H.S.=$(\frac{1}{\sin \theta} - \sin \theta) (\frac{1}{\cos \theta} - \cos \theta) (\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta})$
$= (\frac{1-\sin^2 \theta}{\sin \theta}) (\frac{1-\cos^2 \theta}{\cos \theta}) (\frac{\sin^2 \theta+\cos^2 \theta}{\cos \theta \sin \theta})$
$= (\frac{\cos^2 \theta}{\sin \theta}) \times (\frac{\sin^2 \theta}{\cos \theta}) \times (\frac{1}{\cos \theta \sin \theta})$
$=1 = \text{R.H.S}$

LHS $= \sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}$ ($1/2$)
$= \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}$ ($1/2$)
$= \frac{1}{\sin \theta \cos \theta}$ ($1$)
$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$ ($1/2$)
$= \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta}$
$= \tan \theta + \cot \theta = \text{RHS}$ ($1/2$)

$LHS = \frac{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta}}{\sin \theta \cos \theta}$
$= \frac{\frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}}{\sin \theta \cos \theta}$
$= \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$
$= \frac{\sin^2 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$
$= \frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta}$
$= \sec^2 \theta - \operatorname{cosec}^2 \theta = RHS$

LHS = $\tan^2 A + \frac{1}{\sec^2 A}$
$= \tan^2 A + \cos^2 A$
RHS = $\frac{1}{\cos^2 A} - \sin^2 A$
$= \sec^2 A - \sin^2 A$
$= \tan^2 A + 1 - \sin^2 A$
$= \tan^2 A + \cos^2 A$
$\therefore$ LHS = RHS

LHS $= \frac{\tan^3 \theta}{\sec^2 \theta} + \frac{\cot^3 \theta}{\cosec^2 \theta}$
$= \frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta}$
$= \frac{\sin^4 \theta + \cos^4 \theta}{\sin \theta \cos \theta}$
$= \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta}$
$= \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} - \frac{2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta}$
$= cosec \theta \sec \theta - 2 \sin \theta \cos \theta = \text{RHS}$

$1 + \sin^2 \theta = 3 \sin \theta \cos \theta$
$\Rightarrow (\sin^2 \theta + \cos^2\theta) + \sin^2 \theta – 3 \sin \theta \cos \theta = 0$
$\Rightarrow 2 \sin^2 \theta + \cos^2 \theta – 3 \sin \theta \cos \theta = 0$
Dividing by $\cos^2 \theta$, we get
$2 \tan^2 \theta - 3 \tan \theta + 1 = 0$
$\Rightarrow (2 \tan \theta – 1)( \tan \theta – 1) = 0$
$\therefore \tan \theta = \frac{1}{2}$ or $1$

Given $\sin A + \cos A = \sqrt{3}$
Squaring both sides
$\sin^2A + \cos^2A + 2 \sin A \cos A = 3$
$\Rightarrow \sin A \cos A = 1$
$\frac{1}{\sin A \cos A} = 1$
$\frac{\sin^2A+\cos^2A}{\sin A \cos A} = 1$
$\therefore \tan A + \cot A = 1$

LHS = $(1 + \cot^2\theta)(1 + \tan^2\theta)$
$= \csc^2\theta \cdot \sec^2\theta$
$= \frac{1}{\sin^2\theta \cos^2\theta}$
$= \frac{1}{\sin^2\theta (1-\sin^2\theta)}$
$= \frac{1}{\sin^2\theta-\sin^4\theta} = RHS$

LHS = $\frac{\sqrt{\csc \theta-1} \sqrt{\csc \theta-1} + \sqrt{\csc \theta+1} \sqrt{\csc \theta+1}}{\sqrt{(\csc \theta+1)(\csc \theta-1)}}$
$= \frac{\csc \theta-1 + \csc \theta+1}{\sqrt{\csc^2 \theta-1}}$
$= \frac{2 \csc \theta}{\sqrt{\cot^2 \theta}}$
$= \frac{2 \csc \theta}{\cot \theta}$
$= \frac{2/\sin \theta}{\cos \theta/\sin \theta} = \frac{2}{\cos \theta} = 2 \sec \theta = RHS$

$LHS = m^2 - n^2$
$= (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2$
$= 4 \tan \theta \sin \theta$ ($1$ mark)
$= 4 \sqrt{\tan^2 \theta \sin^2 \theta}$ ($1/2$ mark)
$= 4 \sqrt{\tan^2 \theta (1 - \cos^2 \theta)}$ ($1/2$ mark)
$= 4 \sqrt{\tan^2 \theta - \sin^2 \theta}$ ($1/2$ mark)
$= 4 \sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)}$
$= 4 \sqrt{mn} = RHS$ ($1/2$ mark)

$LHS = \frac{\frac{1}{\tan A} - 1}{2 - (1+\tan^2 A)}$ ($1$ mark)
$= \frac{\frac{1 - \tan A}{\tan A}}{2 - 1 - \tan^2 A}$ ($1/2$ mark)
$= \frac{1 - \tan A}{\tan A(1 - \tan^2 A)}$ ($1/2$ mark)
$= \frac{1}{\tan A (1 + \tan A)}$ ($1/2$ mark)
$= \frac{\cot A}{1 + \tan A} = RHS$ ($1/2$ mark)

$\cot^2 \theta = \text{cosec}^2 \theta - 1 = (x + \frac{1}{4x})^2 - 1$ ($1$)
$= (x - \frac{1}{4x})^2$ ($1/2$)
$\Rightarrow \cot \theta = (x - \frac{1}{4x})$ or $(-x + \frac{1}{4x})$ ($1/2$)
$\text{cosec } \theta + \cot \theta = (x + \frac{1}{4x}) + (x - \frac{1}{4x})$ or $(x + \frac{1}{4x}) + (-x + \frac{1}{4x})$ ($1/2+1/2$)
$= 2x$ or $\frac{1}{2x}$

LHS $$\begin{aligned}& = \left(\frac{1-\cos^2 \text{A}}{\cos \text{A}}\right) \left(\frac{1-\sin^2 \text{A}}{\sin \text{A}}\right) \\ & = \frac{\sin^2 \text{A} \cos^2 \text{A}}{\cos \text{A} \sin \text{A}} \\ & = \sin \text{A} \cdot \cos \text{A} \\ & \text{RHS} = \frac{1}{\frac{\sin \text{A}}{\cos \text{A}} + \frac{\cos \text{A}}{\sin \text{A}}} \\ & = \frac{1}{\frac{\sin^2 \text{A} + \cos^2 \text{A}}{\sin \text{A} \cos \text{A}}} \\ & = \sin \text{A} \cdot \cos \text{A} \\ & \therefore \text{LHS} = \text{RHS}\end{aligned}$$

LHS = $\frac{\cos A+\sin A-1}{\cos A-\sin A+1}$
$= \frac{\cot A+1-\text{cosec } A}{\cot A-1+\text{cosec } A}$
$= \frac{\cot A-\text{cosec } A+\text{cosec}^2A- \cot^2 A}{\cot A-1+\text{cosec } A}$
$= \frac{(\text{cosec } A-\cot A)(-1+\text{cosec } A+\cot A)}{\cot A-1+\text{cosec } A}$
$= \text{cosec } A - \cot A = \text{RHS}$

LHS = $p^2 – q^2$
$= (\cot \theta + \cos \theta)^2 – (\cot \theta – \cos \theta)^2$
$= [(\cot \theta + \cos \theta) + (\cot \theta – \cos \theta)][(\cot \theta + \cos \theta) – (\cot \theta - \cos \theta)]$
$= 2 \cot \theta \times 2 \cos \theta = 4 \cot \theta \cos \theta$
RHS = $4\sqrt{pq}$
$= 4\sqrt{(\cot \theta + \cos \theta) (\cot \theta – \cos \theta)}$
$= 4\sqrt{\cot^2\theta - \cos^2\theta}$
$= 4\sqrt{\cos^2\theta(\text{cosec}^2\theta – 1)}$
$= 4\sqrt{\cos^2\theta \times \cot^2\theta}$
$= 4 \cot \theta \cos \theta$
$\therefore$ LHS = RHS

$LHS = \frac{\cos A + \sin A - 1}{\cos A - \sin A + 1} = \frac{\cot A + 1 - \csc A}{\cot A - 1 + \csc A}$
$= \frac{\cot A - \csc A + \csc^2 A - \cot^2 A}{\cot A - 1 + \csc A}$
$= \frac{(\csc A - \cot A)(-1 + \csc A + \cot A)}{\cot A - 1 + \csc A}$
$= \csc A - \cot A = RHS$

$LHS = \frac{1 + \frac{1}{\sin A}}{\frac{1}{\sin A}} = \sin A + 1$
$= \frac{(\sin A + 1)(1 - \sin A)}{1 - \sin A}$
$= \frac{1 - \sin^2 A}{1 - \sin A}$
$= \frac{\cos^2 A}{1 - \sin A} = RHS$

$LHS = \frac{1 + \frac{1}{\sin A}}{\frac{1}{\sin A}} = \sin A + 1$
$= \frac{(\sin A + 1)(1 - \sin A)}{1 - \sin A} = \frac{1 - \sin^2 A}{1 - \sin A} = \frac{\cos^2 A}{1 - \sin A} = RHS$

$LHS = \frac{\cos \theta - 2\cos^3 \theta}{\sin \theta - 2\sin^3 \theta} + \cot \theta = \frac{\cos \theta(1 - 2\cos^2 \theta)}{\sin \theta(1 - 2\sin^2 \theta)} + \cot \theta$ ($\frac{1}{2}$ mark).
$= \frac{\cos \theta}{\sin \theta} [\frac{\sin^2 \theta + \cos^2 \theta - 2\cos^2 \theta}{\sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta}] + \cot \theta$ (1 mark).
$= \frac{\cot \theta(\sin^2 \theta - \cos^2 \theta)}{(\cos^2 \theta - \sin^2 \theta)} + \cot \theta$ (1 mark).
$$\begin{aligned}& = -\cot \theta + \cot \theta \\ & = 0 = RHS\end{aligned}$$ ($\frac{1}{2}$ mark).

Given: $\sin \theta + \cos \theta = x$.
Squaring both sides
$\sin^2 \theta + \cos^2 \theta + 2\cos \theta \sin \theta = x^2$. $2\sin \theta \cos \theta = x^2 - 1$ (1 mark).
$RHS = \frac{2 - (2\sin \theta \cos \theta)^2}{2} = \frac{2 - 4\sin^2 \theta \cos^2 \theta}{2} = 1 - 2\sin^2 \theta \cos^2 \theta$ ($\frac{1}{2} + \frac{1}{2}$ marks).
$$\begin{aligned}& = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = (\sin^4 \theta + \cos^4 \theta) \\ & = LHS\end{aligned}$$ ($\frac{1}{2} + \frac{1}{2}$ marks).

$x = h + a \cos \theta \Rightarrow \frac{x-h}{a} = \cos \theta$ (I) (1 Mark)
$y = k + b \sin \theta \Rightarrow \frac{y-k}{b} = \sin \theta$ (II) (1 Mark)
$\therefore LHS = (\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = \cos^2\theta + \sin^2\theta = 1 = RHS$ (III) (1 Mark)

LHS $= \frac{\sec^3 \theta}{(\sec^2 \theta - 1)} + \frac{\operatorname{cosec}^3 \theta}{(\operatorname{cosec}^2 \theta - 1)}$ (1 Mark)
$= \frac{\sec^3 \theta}{\tan^2 \theta} + \frac{\operatorname{cosec}^3 \theta}{\cot^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos^3 \theta} \times \frac{\cos^2 \theta}{\sin^2 \theta} + \frac{1}{\sin^3 \theta} \times \frac{\sin^2 \theta}{\cos^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta}$ (1/2 Mark)
$= \frac{1}{\sin \theta \cos \theta} [\frac{1}{\sin \theta} + \frac{1}{\cos \theta}]$ (1/2 Mark)
$= \sec \theta \cdot \operatorname{cosec} \theta (\sec \theta + \operatorname{cosec} \theta) = \text{RHS}$ (1/2 Mark)

LHS $= (p^2 - q^2) \sec^2 \alpha$ (1/2 Mark)
$= (\frac{\sec^2 \alpha}{\operatorname{cosec}^2 \beta} - \frac{\tan^2 \alpha}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1 Mark)
$= (\frac{\sec^2 \alpha - \tan^2 \alpha}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1 Mark)
$= (\frac{1}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1/2 Mark)
$= p^2 = \text{RHS}$

$LHS = \frac{1 + cosec \theta}{cosec \theta} = \frac{1}{cosec \theta} + \frac{cosec \theta}{cosec \theta} = \sin \theta + 1$ (1 Mark)
$RHS = \frac{\cos^2 \theta}{1 - \sin \theta} = \frac{1 - \sin^2 \theta}{1 - \sin \theta}$ (1 Mark)
$= \frac{(1 - \sin \theta)(1 + \sin \theta)}{1 - \sin \theta} = 1 + \sin \theta$ (1 Mark)
$LHS = RHS$

$(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$ (1/2 Mark)
$\Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
$\Rightarrow \sin \theta \cos \theta = 1$ --- (i) (1 Mark)
LHS $= \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2\theta+\cos^2\theta}{\cos \theta \sin \theta}$ (1 Mark)
$= \frac{1}{\cos \theta \sin \theta}$ (1/2 Mark)
$= 1$ [using (i)]
$= \text{RHS}$

LHS $= (\sin A + \frac{1}{\cos A})^2 + (\cos A + \frac{1}{\sin A})^2$ (1/2 Mark)
$= \sin^2 A + \frac{1}{\cos^2 A} + \frac{2 \sin A}{\cos A} + \cos^2 A + \frac{1}{\sin^2 A} + \frac{2 \cos A}{\sin A}$ (1 Mark)
$= 1 + (\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}) + 2 (\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})$
$= 1 + \frac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A} + 2 \frac{\sin^2 A + \cos^2 A}{\cos A \sin A}$ (1 Mark)
$= 1 + \frac{1}{\cos^2 A \sin^2 A} + \frac{2}{\cos A \sin A}$
$= 1 + \sec^2 A cosec^2 A + 2 \sec A cosec A$
$= (1 + \sec A cosec A)^2 = RHS$

L.H.S. $= \frac{\sec^2 x - \tan^2 x}{\sec x - \tan x} - \sec x$ (I) ($\frac{1}{2}$ Mark)
$= \sec x + \tan x - \sec x$
$= \tan x$ (II) (1 Mark)
R.H.S. $= \sec x - \frac{\sec^2 x - \tan^2 x}{\sec x + \tan x}$ (III) ($\frac{1}{2}$ Mark)
$= \sec x - (\sec x - \tan x)$
$= \tan x$ (IV) (1 Mark)
L.H.S. $=$ R.H.S.
Alternate Solution:
Reframing, $\frac{1}{\sec x - \tan x} + \frac{1}{\sec x + \tan x} = \frac{2}{\cos x}$
LHS $= \frac{(\sec x + \tan x) + (\sec x - \tan x)}{(\sec x - \tan x) (\sec x + \tan x)}$ (I) (1 Mark)
$= \frac{2 \sec x}{\sec^2 x - \tan^2 x}$ (II) (1 Mark)
$= 2 \sec x$
$= \frac{2}{\cos x} = \text{RHS}$ (III) ($\frac{1}{2}$ Mark)
(IV) ($\frac{1}{2}$ Mark)

L.H.S. $= \frac{m^2 - 1}{m^2 + 1}$ (I) (1 Mark)
$= \frac{(\sec \theta+\tan \theta)^2 - (\sec^2 \theta - \tan^2 \theta)}{(\sec \theta+\tan \theta)^2 + (\sec^2 \theta - \tan^2 \theta)}$ (II) (1 Mark)
$= \frac{(\sec \theta+\tan \theta) (\sec \theta+\tan \theta - \sec \theta+\tan \theta)}{(\sec \theta+\tan \theta) (\sec \theta+\tan \theta + \sec \theta - \tan \theta)}$ (III) (1/2 Mark)
$= \frac{\tan \theta}{\sec \theta}$ (IV) (1/2 Mark)
$= \sin \theta = \text{R.H.S.}$

$\cos A + \sin A = \sqrt{2} \cos A$ (i) (1 Mark)
Squaring equation (i) both sides to get
$\cos^2 A + \sin^2 A + 2 \sin A \cos A = 2 \cos^2 A$ (1 Mark)
$\Rightarrow 2 \sin A \cos A = \cos^2 A - \sin^2 A$ ($\frac{1}{2}$ Mark)
$\Rightarrow 2 \sin A \cos A = (\cos A + \sin A)(\cos A - \sin A)$ ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{2 \sin A \cos A}{\cos A + \sin A} = (\cos A - \sin A)$ ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{2 \sin A \cos A}{\sqrt{2} \cos A} = (\cos A - \sin A)$ [using (i)] ($\frac{1}{2}$ Mark)
$\Rightarrow (\cos A - \sin A) = \sqrt{2} \sin A$

LHS $= \frac{1 + \frac{\cos^2 A}{\sin^2 A}}{1 + \frac{\sin^2 A}{\cos^2 A}} = \frac{\frac{\sin^2 A + \cos^2 A}{\sin^2 A}}{\frac{\cos^2 A + \sin^2 A}{\cos^2 A}} = \frac{\frac{1}{\sin^2 A}}{\frac{1}{\cos^2 A}} = \frac{\cos^2 A}{\sin^2 A}$
$= \frac{\cos^2 A}{\sin^2 A} (\frac{\sin A - \cos A}{\cos A - \sin A})^2 = (\frac{\sin A - \cos A}{\sin A} \cdot \frac{\cos A}{\cos A - \sin A})^2 = (\frac{1 - \cot A}{1 - \tan A})^2 = RHS$

LHS
$= \sqrt{\frac{\text{cosec } A - 1}{\text{cosec } A + 1} \times \frac{\text{cosec } A - 1}{\text{cosec } A - 1}}$
$= \sqrt{\frac{(\text{cosec } A - 1)^2}{\cot^2 A}} = \frac{\text{cosec } A - 1}{\cot A}$
$= \frac{\text{cosec } A}{\cot A} - \frac{1}{\cot A} = \sec A - \tan A = RHS$

LHS $= (\sin A - \frac{1}{\sin A}) (\cos A - \frac{1}{\cos A}) = \frac{\sin^2 A - 1}{\sin A} \times \frac{\cos^2 A - 1}{\cos A}$
$= \sin A \cos A = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A}$
$= \frac{1}{\tan A + \cot A} = RHS$

LHS: $\frac{\cos^2 \theta + (1 + \sin \theta)^2}{(1 + \sin \theta) \cos \theta}$ [1 mark]
$= \frac{\cos^2 \theta + 1 + \sin^2 \theta + 2 \sin \theta}{(1 + \sin \theta) \cos \theta}$ [1/2 mark]
$= \frac{2 (1 + \sin \theta)}{(1 + \sin \theta) \cos \theta}$ [1 mark]
$= 2 \sec \theta = RHS$ [1/2 mark]

$LHS = \frac{\cos A (1 - 2 \cos^2 A)}{\sin A (2 \sin^2 A - 1)}$ [1/2 mark]
$= \frac{\cos A [1 - 2 (1 - \sin^2 A)]}{\sin A (2 \sin^2 A - 1)}$ [1 mark]
$= \frac{\cos A (- 1 + 2 \sin^2 A)}{\sin A (2 \sin^2 A - 1)}$ [1 mark]
$= \cot A$ [1/2 mark]

$LHS = \frac{\tan \theta}{1 + \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 + \tan \theta}$
$= \frac{\tan^2 \theta}{1 + \tan \theta} + \frac{1}{\tan \theta(1 + \tan \theta)}$
$= \frac{1 + \tan^3 \theta}{\tan \theta(1 + \tan \theta)}$
$= \frac{(1 + \tan \theta)(1 + \tan^2 \theta - \tan \theta)}{\tan \theta(1 + \tan \theta)}$
$= \cot \theta + \tan \theta - 1 = RHS$

$LHS = \frac{\cos \theta}{1 - \frac{\sin \theta}{\cos \theta}} + \frac{\sin \theta}{1 - \frac{\cos \theta}{\sin \theta}} = \frac{\cos^2 \theta}{\cos \theta - \sin \theta} - \frac{\sin^2 \theta}{\cos \theta - \sin \theta} = \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\cos \theta - \sin \theta} = \cos \theta + \sin \theta = RHS$

$LHS = \sin^2 \theta + \sec^2 \theta + 2 \sin \theta \sec \theta + \cos^2 \theta + \text{cosec}^2 \theta + 2 \cos \theta \text{cosec } \theta$
$= 1 + (\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}) + 2 (\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta})$
$= 1 + \frac{1}{\sin^2 \theta \cos^2 \theta} + \frac{2}{\sin \theta \cos \theta}$
$= 1 + \sec^2 \theta \text{cosec}^2 \theta + 2 \sec \theta \text{cosec } \theta$
$= (1 + \sec \theta \text{cosec } \theta)^2 = RHS$

$\frac{2}{3} [(\frac{\sqrt{3}}{2})^4 - (\frac{1}{\sqrt{2}})^4] - 3 [(\frac{\sqrt{3}}{2})^2 - (\sqrt{2})^2] + \frac{1}{4}(\sqrt{3})^2 = \frac{113}{24}$

$$\begin{aligned}& LHS = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{\sin^2 A - \cos^2 A} \\ & = \frac{2(\sin^2 A + \cos^2 A)}{1 - \cos^2 A - \cos^2 A} \\ & = \frac{2}{1 - 2\cos^2 A} \\ & = RHS\end{aligned}$$

(i) $\cot x = \frac{\cos x}{\sin x} = \frac{\sqrt{1 - \sin^2 x}}{\sin x}$ (1 Mark)
$= \frac{\sqrt{1-p^2}}{p}$ (1/2 Mark)
(ii) $\frac{1+\tan^2 x}{1+\cot^2 x} = \frac{\sec^2 x}{\operatorname{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x}$ (1/2 + 1/2 Mark)
$= \frac{p^2}{1-p^2}$ (1/2 Mark)

LHS $= \sin^2 A +\operatorname{cosec}^2 A - 2\sin A.\operatorname{cosec} A + \cos^2 A + \sec^2 A - 2\cos A.\sec A$ (1 Mark)
$= (\sin^2 A + \cos^2 A) + 1 + \cot^2 A + 1 + \tan^2 A - 2 - 2$ (1 Mark)
$= 1 + \cot^2 A + \tan^2 A +2-4$ (1/2 Mark)
$= \tan^2 A + \cot^2 A - 1 = \text{RHS}$ (1/2 Mark)

(a) LHS = $\frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A}$ (1 Mark)
= $\frac{\cos A (\frac{1}{\sin A} - 1)}{\cos A (\frac{1}{\sin A} + 1)}$
= $\frac{1 - \sin A}{1 + \sin A}$ (1/2 Mark)
= $\frac{\frac{1}{\cos A} - \frac{\sin A}{\cos A}}{\frac{1}{\cos A} + \frac{\sin A}{\cos A}}$ (1 Mark)
= $\frac{\sec A - \tan A}{\sec A + \tan A}$ = RHS (1/2 Mark)

(a) LHS = $\frac{\sec^2\theta + \tan^2\theta + 2\sec\theta \tan\theta - 1}{\sec^2\theta + \tan^2\theta + 2\sec\theta \tan\theta + 1}$ (1)
$= \frac{2\tan^2\theta + 2\sec\theta \tan\theta}{2\sec^2\theta + 2\sec\theta \tan\theta}$ (1)
$= \frac{2\tan\theta (\tan\theta + \sec\theta)}{2\sec\theta (\sec\theta + \tan\theta)}$ (½)
$= \sin\theta = RHS$. (½)

LHS = $(\tan^2 \theta + 1) + (\cot^2 \theta + 1)$ (1 Mark)
$= \sec^2 \theta + cosec^2 \theta$ (1/2 Mark)
$= \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}$ (1/2 Mark)
$= \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos^2 \theta \sin^2 \theta}$ (1/2 Mark)
$= \sec^2 \theta cosec^2 \theta = \text{RHS}$

LHS = $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \sqrt{\frac{\frac{1}{\sec A} - 1}{\frac{1}{\sec A} + 1}}$ (1 Mark)
$= \sqrt{\frac{\sec A - 1}{\sec A + 1}}$ (1 Mark)
$= \sqrt{\frac{\sec A - 1}{\sec A + 1} \times \frac{\sec A + 1}{\sec A + 1}} = \sqrt{\frac{\sec^2 A - 1}{(\sec A + 1)^2}} = \frac{\tan A}{\sec A + 1} = \text{RHS}$ (1 Mark)

$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$
$\sin \theta = (\sqrt{2} – 1) \cos \theta$
$= \frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1} \cos \theta$ (1.5 Mark)
$\Rightarrow (\sqrt{2} + 1) \sin \theta = \cos \theta$ (1 Mark)
$\Rightarrow \sqrt{2} \sin \theta = \cos \theta – \sin \theta$ (0.5 Mark)

(i) LHS $= \sqrt{1 + \tan^2\theta + 1 + \cot^2\theta}$
$= \sqrt{\tan^2\theta + \cot^2\theta + 2 \times \tan\theta \times \cot\theta}$
$= \sqrt{(\tan\theta + \cot\theta)^2}$
$= \tan\theta + \cot\theta = \text{RHS}$
(ii) $\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$
$= \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1+\sqrt{3})}$
$= \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} \times \frac{\sqrt{2}(1-\sqrt{3})}{\sqrt{2}(1-\sqrt{3})}$
$= \frac{\sqrt{6}(1-\sqrt{3})}{4(1-3)} = \frac{\sqrt{6}-\sqrt{18}}{-8} = \frac{3\sqrt{2}-\sqrt{6}}{8}$

Given, $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta) + y \cos \theta (\cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta) + x \sin \theta (\cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x \sin \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta$
$\Rightarrow x = \cos \theta$
Given, $x \sin \theta = y \cos \theta$
$\Rightarrow \cos \theta \sin \theta = y \cos \theta$
$\Rightarrow y = \sin \theta$
LHS $= x^2 + y^2 = (\cos \theta)^2 + (\sin \theta)^2 = 1 = \text{RHS}$

LHS $= \sin^6\theta+ \cos^6 \theta$
$= (\sin^2\theta)^3 +(\cos^2\theta)^3$ ($\frac{1}{2}$)
$= (\sin^2\theta+ \cos^2\theta)[(\sin^2\theta)^2 + (\cos^2\theta)^2 - \sin^2\theta\cos^2\theta]$ (1)
$= \sin^4\theta+ \cos^4\theta-\sin^2\theta\cos^2\theta$
$= (\sin^2\theta+ \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta - \sin^2\theta\cos^2\theta$ (1)
$= 1 - 3 \sin^2\theta\cos^2\theta$ (1)
$= RHS$ ($\frac{1}{2}$)

LHS $= \frac{(1+\sin\theta)^2-(1-\sin \theta)^2}{(1+\sin \theta) (1-\sin \theta)}$ (2 Marks)
$= \frac{4 \sin \theta}{1-\sin^2\theta}$ (1 Mark)
$= \frac{4 \sin \theta}{\cos^2\theta}$ (1 Mark)
$= 4 \tan \theta \sec \theta = \text{RHS}$ (1 Mark)

(A) $40$ m