Question 1: Evaluate  each of the following:

i) $cosec \ 30^o + \cot 45^o = 2 + 1 = 3$

ii) $\cos 30^o \cos 45^o - \sin 30^o \sin 45^o =$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{1}{\sqrt{2}}$ $- \frac{1}{2}$ $\times$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{\sqrt{3} -1}{2\sqrt{2}}$

iii) $\tan 30^o \sec 45^o + \tan 60^o \sec 30^o =$ $\frac{1}{\sqrt{3}}$ $\times \sqrt{2} + \sqrt{3} \times$ $\frac{2}{\sqrt{3}}$ $=$ $\frac{\sqrt{2} + 2\sqrt{3}}{\sqrt{3}}$

iv) $\sin 30^o \sin 45^o + \cos 30^o \cos 45^o =$ $\frac{1}{2}$ $\times$ $\frac{1}{\sqrt{2}}$ $+$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{\sqrt{3}+1}{2\sqrt{2}}$

v) $\frac{\sin^2 45^o + \cos^2 45^o}{\tan^2 60^o}$ $=$ $\{ \frac{ (\frac{1}{\sqrt{2}} )^2+ ( \frac{1}{\sqrt{2}} )^2}{(\sqrt{3})^2} \}$ $=$ $\frac{\frac{1}{2}+\frac{1}{2}}{3}$ $=$ $\frac{1}{3}$

vi) $\frac{\sin 30^o - \sin 90^o \ 2 \cos 0^o}{ \tan 30^o \tan 60^o}$ $=$ $\{ \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}} \}$ $=$ $\frac{3}{2}$

vii) $\frac{\sin 60^o}{\cos^2 45^o}$ $- \cot 30^o + 15 \cos 90^o =$ $\{ \frac{\frac{\sqrt{3}}{2}}{( \frac{1}{\sqrt{2}} )^2} \}$ $- \sqrt{3} + 15 \times 0 = \sqrt{3} - \sqrt{3} = 0$

viii) $\frac{5 \sin^2 30^o + \cos^2 45^o - 4 \tan^2 30^o}{2 \sin 30^o \cos 30^o + \tan 45^o}$ $=$ $\{ \frac{5 \times (\frac{1}{2} )^2 + ( \frac{1}{\sqrt{2}} )^2 - 4 ( \frac{1}{\sqrt{3}} )^2}{ 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} + 1} \}$

$=$ $\frac{\frac{5}{4} + \frac{1}{2} - \frac{4}{3}}{\frac{\sqrt{3}}{2} + 1}$ $=$ $\frac{5}{6}$ $(2 - \sqrt{3})$

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Question 2: Find the value of \theta in each of the following:

i) $2 \sin \theta =\sqrt{3}$

$\Rightarrow \sin 2 \theta =$ $\frac{\sqrt{3}}{2}$ $\Rightarrow \sin 2 \theta = \sin 60^o$

$\Rightarrow 2 \theta = 60^o$ $\Rightarrow \theta = 30^o$

ii) $2 \cos 3 \theta = 1$

$\Rightarrow \cos 3\theta = \cos 60^o$ $\Rightarrow 3 \theta = 60^o$

$\Rightarrow \theta = 20^o$

iii) $\sqrt{3} \tan 2\theta - 3 = 0$

$\Rightarrow \tan 2 \theta =$ $\frac{3}{\sqrt{3}}$ $= \sqrt{3}$ $\Rightarrow \tan 3 \theta = \tan 60^o$

$\Rightarrow 2 \theta = 60^o$ $\Rightarrow \theta = 30^o$

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Question 3: Evaluate / Prove the following:

i) $\frac{\cos 37^o}{\sin 53^o}$ $=$ $\frac{\cos (90^o-53^o)}{\sin 53^o}$ $=$ $\frac{\sin 53^o}{\sin 53^o}$ $=$ $1$

ii) $\frac{\tan 54^o}{\cot 36^o}$ $=$ $\frac{\tan (90^o-36^o)}{\cot 36^o}$ $=$ $\frac{\cot 36^o}{\cot 36^o}$ $=$ $1$

iii) $\sin 39^o - \cos 51^o = \sin 39^o - \cos (90^o-39^o) = \sin 39^o - \sin 39^o = 0$

iv) $\cot 34^o - \tan 56^o = \cot 34^o - \tan (90^o-34^o) = \cot 34^o - \cot 34^o = 0$

v) $\frac{\cos 80^o}{\sin 10^o}$ $+$ $\cos 59^o \ cosec \ 31^o =$ $\frac{\cos (90^o-10^o)}{\sin 10^o}$ $+$ $\cos 59^o \ cosec \ (90^o-59^o)$

$=$ $\frac{\sin 10^o}{\sin 10^o}$ $+$ $\cos 59^o \sec 59^o = 1 + 1 = 2$

vi) $\sec 50^o \sin 40^o + \cos 40^o \ cosec \ 50^o$

$= \sec 50^o \sin (90^o - 50^o) + \cos 40^o \ cosec \ (90^o-40^o)$

$= \sec 50^o \cos 50^o + \cos 40^o \sec 40^o = 1 + 1 = 2$

vii) $\Big($ $\frac{\sin 35^o}{\cos 55^o}$ $\Big)^2 + \Big($ $\frac{\cos 55^o}{\sin 35^o}$ $\Big)^2 - 2 \cos 60^o$

$= \Big($ $\frac{\sin (90^o-55^o)}{\cos 55^o}$ $\Big)^2 + \Big($ $\frac{\cos (90^o-35^o)}{\sin 35^o}$ $\Big)^2 - 2 \cos 60^o$

$= \Big($ $\frac{\cos 55^o}{\cos 55^o}$ $\Big)^2 + \Big($ $\frac{\sin 35^o}{\sin 35^o}$ $\Big)^2 - 2 \times$ $\frac{1}{2}$

$= 1 + 1 - 1 = 1$

vii)  $\cos (40^o - \theta ) - \sin ( 50^o + \theta) +$ $\frac{\cos^2 40^o + \cos^2 50^o}{\sin^2 40^o + sin^2 50^o}$

$= \sin (90 - (40^o - \theta) ) - \sin ( 50^o + \theta) +$ $\frac{\cos^2 40^o + \cos^2 (90-40^o)}{\sin^2 40^o + sin^2 (90-40^o)}$

$= \sin (50^o + \theta ) - \sin ( 50^o + \theta) +$ $\frac{\cos^2 40^o + \sin^2 40^o}{\sin^2 40^o + \cos^2 40^o}$

$= 0 + 1 = 1$

viii) $\cot 12^o \cot 38^o \cot 52^o \cot 60^o \cot 78^o$

$= \cot (90^o-78^o) \cot (90^o-52^o) \cot 52^o \cot 60^o \cot 78^o$

$= \tan 78^o \tan 52^o \cot 52^o \cot 60^o \cot 78^o$

$= \cot 60^o =$ $\frac{1}{\sqrt{3}}$

ix)  $\tan 5^o \tan 25^o \tan 30^o \tan 65^o \tan 85^o$

$= \tan (90^o-85^o) \tan (90^o-65^o) \tan 30^o \tan 65^o \tan 85^o$

$= \cot 85^o \cot 65^o \tan 30^o \tan 65^o \tan 85^o$

$= \tan 30^o =$ $\frac{1}{\sqrt{3}}$

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Question 4: Express each of the following in terms of trigonometric ratios of angles between $0$ and $45^o$.

i) $\sin 85^o + \ cosec \ 85^o = \sin (90^o-5^o) + \ cosec \ (90^o-5^o) = \cos 5^o + \sec 5^o$

ii) $cosec \ 69^o + \cot 69^o = \ cosec \ (90^o-21^o) + \cot (90^o-21^o) = \sec 21^o + \tan 21^o$

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Question 5: Prove that:

i) $\tan 1^o \tan 2^o \tan 3^o \cdots \tan 89^o = 1$

ii) $\cos 1^o \cos 2^o \cos 3^o \cdots \cos 180^o = 0$

i) LHS $= \tan 1^o \tan 2^o \tan 3^o \cdots \tan 89^o$

$= \tan (90^o-89^o) \tan (90^o-88^o) \tan (90^o-87^o) \cdots \tan 87^o \tan 88^o \tan 89^o$

$= \cot 89^o \cot 88^o \cot 87^o \cdots \tan 87^o \tan 88^o \tan 89^o$

$= 1 \times 1 \times 1 \cdots \times 1 = 1 = RHS$

Hence Proved.

ii) LHS $= \cos 1^o \cos 2^o \cos 3^o \cdots \cos 180^o$

$= \cos 1^o \cos 2^o \cos 3^o \cdots \cos 89^o \cos 90^o \cos 91^o \cdots \cos 179^o \cos 180^o$

$= \cos 1^o \cos 2^o \cos 3^o \cdots \cos 89^o (0) \cos 91^o \cdots \cos 179^o \cos 180^o$

$= 0 =$ RHS.

Hence Proved.

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Question 6: If $A + B = 90^o$, prove that:

$\sqrt{ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} }$ $= \tan A$

Given $A + B = 90^o \Rightarrow B = 90^o - A$

LHS $=$ $\sqrt{ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} }$

$=$ $\sqrt{ \frac{\tan A \tan (90^o - A) + \tan A \cot (90^o - A)}{\sin A \sec (90^o - A)} - \frac{\sin^2 (90^o - A)}{\cos^2 A} }$

$=$ $\sqrt{ \frac{\tan A \cot A + \tan^2 A }{\sin A \ cosec A} - \frac{\cos^2 A}{\cos^2 A} }$

$= \sqrt{1 + \tan^2 A - 1} = \sqrt{\tan^2 A} = \tan A =$ RHS

Hence Proved.

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Question 7: If $A, \ B$ and $C$ are interior angles of triangle ABC, prove that

$\tan \big($ $\frac{B+C}{2}$ $\Big) = \cot \Big($ $\frac{A}{2}$ $\Big)$

Given $A + B + C = 180^o$

$\Rightarrow B + C = 180^o - A$

$\Rightarrow \big($ $\frac{B+C}{2}$ $\Big) = \Big( 90^o -$ $\frac{A}{2}$ $\Big)$

$\Rightarrow \tan \big($ $\frac{B+C}{2}$ $\Big) = \tan \Big( 90^o -$ $\frac{A}{2}$ $\Big)$

$\Rightarrow \tan \big($ $\frac{B+C}{2}$ $\Big) = \cot \Big($ $\frac{A}{2}$ $\Big)$

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Question 8: Find $\theta$ if  $\sin (\theta + 36^o) = \cos \theta$ when $\theta + 36^o$ is an acute angle.

Given $\sin (\theta + 36^o) = \cos \theta$

$\Rightarrow \cos [90^o - (\theta + 36^o) ] = \cos \theta$

$\Rightarrow [ (\theta + 36^o) ] = \theta$

$\Rightarrow 2 \theta = 54^o$

$\Rightarrow \theta = 27^o$

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Question 9: If $\tan 2 \theta = \cot (\theta + 6^o)$, when $2 \theta$ and $\theta+6^o$ are acute angles, find the value of  $\theta$.

Given $\tan 2 \theta = \cot (\theta + 6^o)$

$\Rightarrow \cot (90^o- 2 \theta) = \cot (\theta+6^o)$

$\Rightarrow 90^o - 2 \theta = \theta + 6^o$

$\Rightarrow 84^o = 3 \theta$

$\Rightarrow \theta = 28^o$

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Question 10: If $A, B$ and $C$ are interior angles of $\triangle ABC$, show that

i) $\sin \Big($ $\frac{B+C}{2}$ $\Big) = \cos \Big($ $\frac{A}{2}$ $\Big)$   ii) $\cos \Big($ $\frac{B+C}{2}$ $\Big) = \sin \Big($ $\frac{A}{2}$ $\Big)$

i) Given $A + B + C = 180^o$

$\Rightarrow B + C = 180^o - A$

$\Rightarrow \Big($ $\frac{B+C}{2}$ $\Big) = \Big( 90^o -$ $\frac{A}{2}$ $\Big)$

$\Rightarrow \sin \Big($ $\frac{B+C}{2}$ $\Big) = \sin \Big(90^o -$ $\frac{A}{2}$ $\Big)$

$\Rightarrow \sin \Big($ $\frac{B+C}{2}$ $\Big) = \cos \Big($ $\frac{A}{2}$ $\Big)$

ii) Similarly,

$\Rightarrow \cos \Big($ $\frac{B+C}{2}$ $\Big) = \cos \Big( 90^o -$ $\frac{A}{2}$ $\Big)$

$\Rightarrow \cos \Big($ $\frac{B+C}{2}$ $\Big) = \sin \Big($ $\frac{A}{2}$ $\Big)$

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Question 11: Find the value of $\theta$ for

i) $\cos 2\theta = \sin 4\theta$ where $2\theta, 4\theta < 90^o$

ii) $\sin 3\theta = \cos (\theta - 6^o)$ where $3\theta, (\theta-6^o) < 90^o$

i) $\cos 2\theta = \sin 4\theta$

$\Rightarrow \cos 2\theta = \cos (90^o - 4\theta)$

$\Rightarrow 2\theta = 90^o- 4\theta$

$\Rightarrow 6\theta = 90^o$

$\Rightarrow \theta = 15^o$

ii) $\sin 3\theta = \cos (\theta - 6^o)$

$\Rightarrow \sin 3\theta = \sin [90^o - (\theta-6)]$

$\Rightarrow 3\theta = 90^o - (\theta - 6^o)$

$\Rightarrow 3\theta = 90^o - \theta +6^o$

$\Rightarrow 4\theta = 96^o \Rightarrow \theta = 19^o$

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Question 12: Prove the following:

i) $\tan 20^o \tan 35^o \tan 45^o \tan 55^o \tan 70^o = 1$

ii) $\frac{\cos 80^o}{\sin 10^o}$ $+ \cos 59^o \ cosec \ 31^o = 2$

iii) $\frac{\cos (90^o - A) \sec (90^o-A) \tan A}{cosec \ (90^o - A) \sin (90^o - A) \cot (90^o- A)}$ $+$ $\frac{\tan (90^o-A)}{\cot A}$ $= 2$

i) LHS $= \tan 20^o \tan 35^o \tan 45^o \tan 55^o \tan 70^o$

$= \tan (90^o-70^o) \tan (90^o-55^o) \tan 45^o \tan 55^o \tan 70^o$

$= \cot 70^o \cot 55^o \tan 45^o \tan 55^o \tan 70^o$

$= \tan 45^o = 1 =$ RHS

Hence Proved.

ii) LHS $=$ $\frac{\cos 80^o}{\sin 10^o}$ $+ \cos 59^o \ cosec \ 31^o$

$=$ $\frac{\cos (90^o-10^o)}{\sin 10^o}$ $+ \cos 59^o \ cosec \ (90^o- 59^o)$

$=$ $\frac{\sin 10^o}{\sin 10^o}$ $+ \cos 59^o \sec 59^o$

$= 1 + 1 = 2 =$ RHS

Hence Proved.

iii) LHS $=$ $\frac{\cos (90^o - A) \sec (90^o-A) \tan A}{cosec \ (90^o - A) \sin (90^o - A) \cot (90^o- A)}$ $+$ $\frac{\tan (90^o-A)}{\cot A}$

$=$ $\frac{\sin A \ cosec \ A \tan A}{\sec A \cos A \tan A}$ $+$ $\frac{\cot A}{ \cot A}$

$= 1 + 1 = 2$

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Question 13: What is the max value of i) $\frac{1}{\sec A}$ and ii) $\frac{1}{cosec \ A}$

i) $\frac{1}{\sec A}$ $= \cos A \Rightarrow max \ value = 1$

ii) $\frac{1}{cosec \ A}$ $= \sin A \Rightarrow max \ value = 1$

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Question 14: If $\tan A =$ $\frac{4}{5}$, find the value of $\frac{\cos A - \sin A}{\cos A + \sin A}$

$\frac{\cos A - \sin A}{\cos A + \sin A}$ $=$ $\frac{1 - \tan A}{1 + \tan A}$ $=$ $\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}$ $=$ $\frac{1}{9}$

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Question 15: If $\tan A =$ $\frac{1}{\sqrt{5}}$, what is the value of $\frac{cosec^2 \ A - \sec^2 A}{cosec^2 \ A + \sec^2 A}$

$\frac{cosec^2 \ A - \sec^2 A}{cosec^2 \ A + \sec^2 A}$

$=$ $\frac{\frac{1}{\sin^2 A} - \frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A} + \frac{1}{\cos^2 A}}$

$=$ $\frac{\cos^2 A - \sin^2 A}{\cos^2 A + \sin^2 A}$

$=$ $\frac{1 - \tan^2 A}{1 + \tan^2 A}$

$=$ $\frac{1 - (\frac{1}{\sqrt{5}})^2}{1 + (\frac{1}{\sqrt{5}})^2}$

$=$ $\frac{4}{6}$ $=$ $\frac{2}{3}$

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