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) $3$

(D) $0$

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

$BC = \sqrt{(-8 - 1)^2 + (7 - 1)^2} = \sqrt{117} = 3\sqrt{13}$ (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)

$\tan \theta = \frac{1}{\sqrt{7}}$
$\Rightarrow \sec^2\theta = \frac{8}{7}$ 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) $\frac{\tan^2 A - \sec^2 A}{\cot^2 A - \csc^2 A} = \frac{(\frac{15}{8})^2 - (\frac{17}{8})^2}{(\frac{8}{15})^2 - (\frac{17}{15})^2} = 1$

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

(C) 0

(B) $\cos 60^\circ$

(C) 2:1

(C) $3$

(A) $-1$

(C) $1$

(D) $-\sqrt{3}$

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

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

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

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

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

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

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

$\frac{(\frac{\sqrt{3}}{2})^3 - \frac{1}{\sqrt{3}}}{(\frac{1}{\sqrt{2}})^2}$ (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)

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

(C) $1$

(d) $30^\circ$

(D) $0^{\circ}$

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

$\tan(A + 2B) = \sqrt{3} \Rightarrow A + 2B = 60^\circ$ (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)

$\sin (A + 2B) = 1 \Rightarrow A + 2B = 90^\circ$ --------(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)

(A) $\cot \theta$

(A) $\tan^2 A$

(B) $\sec^2 A$

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

$\sin A = \frac{1}{2} \Rightarrow A = 30^\circ$, $\tan B = \sqrt{3} \Rightarrow B = 60^\circ$ (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 $(\sec \theta – \tan \theta)$
$= (\tan \theta + \sec \theta) \times \frac{(\sec \theta - \tan \theta)}{(\sec \theta - \tan \theta)}$
$= \frac{(\sec^2\theta-\tan^2\theta)}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = \text{RHS}$

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

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

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

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

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

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