Question 1: $\frac{\sec A -1 }{\sec A + 1} = \frac{1- \cos A }{1+ \cos A}$   [2007]

LHS $=$ $\frac{\sec A -1 }{\sec A + 1}$

$=$ $\frac{\frac{1}{\cos A} -1 }{\frac{1}{\cos A} + 1}$

$=$ $\frac{1- \cos A }{1+ \cos A} =$ RHS. Hence Proved.

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Question 2: $(1-\tan A)^2 + (1 + \tan A)^2 = 2 \ \sec^2 A$   [2005]

LHS $=$ $(1-\tan A)^2 + (1 + \tan A)^2$

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

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

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

$=$ $\frac{2}{\cos^2 A}$ $= 2 \ \sec^2 A =$ RHS. Hence Proved.

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Question 3:   $\frac{\sin A}{1 + \cos A}$ $= cosec A - \cot A$   [2008]

RHS $=$ $cosec A - \cot A$

$=$ $\frac{1}{\sin A} - \frac{\cos A}{\sin A}$

$=$ $\frac{1- \cos A}{\sin A}$

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

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

$=$ $\frac{\sin A}{1 + \cos A} =$ LHS. Hence proved.

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Question 4:   $\sqrt{\frac{1 - \cos A}{1 + \cos A}}$ $= cosec \ A - \cot A$    [2000]

LHS $=$ $\sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$=$ $\sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}}$

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

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

$=$ $\frac{1- \cos A}{\sin A}$

$=$ $cosec A - \cot A =$ RHS. Hence proved.

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Question 5:   $\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A}$     [2013]

LHS $=$ $\sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$=$ $\sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}}$

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

$=$ $\frac{\sin A}{1 + \cos A}$ = RHS. Hence proved.

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Question 6:   $1 -$ $\frac{\cos^2 A}{1 + \sin A}$ $= \sin A$    [2001]

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

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

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

$=$ $\frac{\sin A(1 + \sin A)}{1 + \sin A}$

$=$ $\sin A =$ RHS. Hence proved.

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Question 7:   $\frac{1}{\sin A + \cos A} + \frac{1}{\sin A + \cos A} = \frac{2 \ \sin A}{1 - 2 \ \cos^2 A}$    [2002]

LHS $=$ $\frac{1}{\sin A + \cos A} + \frac{1}{\sin A + \cos A}$

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

$=$ $\frac{2 \ \sin A}{1 - 2 \ \cos^2 A}$ = RHS. Hence proved.

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Question 8:   $\frac{\sin \theta . \tan \theta}{1 - \cos \theta}$ $= 1 + \sec \theta$    [2006]

LHS $=$ $\frac{\sin \theta . \tan \theta}{1 - \cos \theta}$

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

$=$ $\frac{(1-\cos \theta)(1 + \cos \theta)}{\cos \theta (1 - \cos \theta)}$

$=$ $\frac{1 + \cos \theta}{\cos \theta}$

$=$ $\sec \theta + 1$ = RHS. Hence proved.

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Question 9:   $\frac{2 \ \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o}$   [2006]

$\frac{2 \ \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o}$

$=$ $\frac{2 \ \tan (90^o - 37^o)}{\cot 37^o}-\frac{\cot (90^o - 10^o)}{\tan 10^o}$

$=$ $\frac{2 \ \cot 37^o}{\cot 37^o}-\frac{\tan 10^o}{\tan 10^o}$ $= 0$

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Question 10:   $\cos^2 26^o + \cos 64^o.\sin 26^o +$ $\frac{\tan 36^o}{\cot 54^o}$    [2012]

$\cos^2 26^o + \cos 64^o.\sin 26^o +$ $\frac{\tan 36^o}{\cot 54^o}$

= $\cos^2 26^o + \cos (90^o - 26^o).\sin 26^o +$ $\frac{\tan (90^o - 54^o)}{\cot 54^o}$

= $\cos^2 26^o + \sin^2 26^o +$ $\frac{\cot 54^o}{\cot 54^o}$

$= 2$

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Question 11:   $3 \ \cos 80^o . cosec \ 10^o + 2 \sin 59^o.\sec 31^o$    [2013]

$3 \cos 80^o . cosec \ 10^o + 2 \sin 59^o.\sec 31^o$

$= 3 \cos 80^o . cosec (90^o - 80^o) + 2 \sin 59^o.\sec (90^o - 59^o)$

$= 3 \cos 80^o . \sec 80^o + 2 \sin 59^o . cosec \ 59^o$

$= 3 + 2 = 5$

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Question 12:   $\frac{\sin 80^o}{\cos 10^o}$ $+ \sin 59^o . \sec 31^o$    [2007]

$\frac{\sin 80^o}{\cos 10^o}$ $+ \sin 59^o . \sec 31^o$

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

$=$ $\frac{\cos 10^o}{\cos 10^o}$ $+ \sin 59^o . cosec \ 59^o$ $= 2$

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Question 13: $14 \ \sin 30^o + 6 \cos 60^o - 5 \ \tan 45^o$    [2004]

$14 \ \sin 30^o + 6 \cos 60^o - 5 \ \tan 45^o$

$= 14 \ \sin (90^o - 60^o) + 6 \cos 60^o - 5 \ \tan 45^o$

$= 14 \ \cos 60^o + 6 \cos 60^o - 5 \ \tan 45^o$

$= 20 \cos 60^o - 5 \ \tan 45^o$

$= 20 \times \frac{1}{2} - 5 \times 1$

$= 10-5 = 5$

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Question 14: Evaluate without using trigonometric tables:

$2$ $(\frac{\tan 35^o}{\cot 55^o})^2$ $+$ $(\frac{\cot 55^o}{\tan 35^o})^2$ $- 3$ $(\frac{\sec 40^o}{cosec 50^o})$    [2011]

$2$ $(\frac{\tan 35^o}{\cot 55^o})^2$ $+$ $(\frac{\cot 55^o}{\tan 35^o})^2$ $- 3$ $(\frac{\sec 40^o}{cosec 50^o})$

$= 2$ $(\frac{\tan 35^o}{\cot (90^o - 35^o)})^2$ $+$ $(\frac{\cot 55^o}{\tan (90^o - 55^o)})^2$ $- 3$ $(\frac{sec 40^o}{cosec \ (90^o - 40^o)})$

$= 2$ $(\frac{\tan 35^o}{\tan 35^o})^2$ $+$ $(\frac{\cot 55^o}{\cot 55^o})^2$ $- 3$ $(\frac{\sec 40^o}{\sec 40^o})$

$= 2 + 1 - 3 = 0$

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Question 15: Prove that $(cosec\ A- \sin A) (\sec A-\cos A) \ \sec^2 A = \tan A$    [2011]

$(cosec \ A- \sin A) (\sec A-\cos A) \ \sec^2 A = \tan A$

LHS $= (cosec \ A- \sin A) (\sec A-\cos A) \ \sec^2 A$

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

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

$= 2$ $\frac{\sin A}{\cos A}$ $=\tan A =$ RHS

Hence Proved.

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Question 16: Without using trigonometric tables, evaluate

$\sin^2 34^o + \sin^2 56^o + 2 \ \tan 18^o \tan 72^o - \cot ^2 30^o$   [2014]

Given $\sin^2 34^o + \sin^2 56^o + 2 \ \tan 18^o \tan 72^o - \cot ^2 30^o$

$= \sin^2 34^o + \sin^2 (90^o - 34^o) + 2 \ \tan 18^o \ \tan (90^o-18^o) - \cot ^2 30^o$

$= \sin^2 34^o + \cos^2 34^o + 2 \ \tan 18^o \ \cot 18^o - (\sqrt{3})^2$

$= 1+ 2 \ \tan 18^o \times \frac{1}{\tan 18^o}-3$

$= 1+2-3 = 0$

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Question 17:  Prove the identity:

$(\sin \theta + \cos \theta) (\tan \theta + \cot \theta) = \sec \theta + cosec \theta$    [2014]

LHS $= (\sin \theta + \cos \theta)(\tan \theta+ \cot \theta)$

$= (\sin \theta + \cos \theta)$ $(\frac{\sin \theta}{\cos \theta} + \frac{cos \theta}{\sin \theta})$

$= (\sin \theta + \cos \theta)$ $(\frac{\sin^2 \theta+ \cos^2 \theta}{\sin \theta . \cos \theta})$

$= (\sin \theta + \cos \theta)$ $(\frac{1}{\sin \theta . \cos \theta})$

$=$ $(\frac{\sin \theta}{\sin \theta . \cos \theta}) + (\frac{\cos \theta}{sin \theta . \cos \theta})$

$=$ $\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$

$= \sec \theta+cosec \ \theta = RHS$

Hence Proved.

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Question 18: If $2 \sin A - 1 = 0$, show that $\sin 3A = 2 \sin A - 4 \sin^3 A$    [2001]

$2 \sin A - 1 = 0$

$\Rightarrow \sin A = \frac{1}{2} \Rightarrow A = 30^o$

Therefore to prove: $\sin 3A = 2 \sin A - 4 \sin^3 A$

LHS $= \sin 3A = \sin 90^o = 1$

RHS $= 2 \sin A - 4 \sin^3 A = 2 \sin 30^o - 4 \sin^3 30^o = \frac{3}{2} - \frac{1}{2} = 1$

Therefore LHS = RHS. Hence proved.

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Question 19: Evaluate: $\frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{cosec 58^o}$ [2000]

$\frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{cosec 58^o}$

$=$ $\frac{3 \sin 72^o}{\cos (90^o - 72^o)}- \frac{\sec 32^o}{cosec (90^o - 32^o)}$

$=$ $\frac{3 \sin 72^o}{\sin 72^o}- \frac{\sec 32^o}{\sec 32^o}$

$= 3 - 1 = 2$

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Question 20: Evaluate: $3 \cos 80^o cosec \ 10^o + 2 \cos 59^o cosec \ 31^o$    [2002]

$3 \cos 80^o cosec \ 10^o + 2 \cos 59^o cosec \ 31^o$

$= 3 \cos 80^o cosec \ (90^o - 80^o) + 2 \cos 59^o cosec \ (90^o - 59^o)$

$= 3 \cos 80^o \sec 80^o + 2 \cos 59^o \sec 59^o$

$= 3 + 2 = 5$

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Question 21: $\frac{\cos 75^o}{\sin 15^o} + \frac{\sin 12^o}{\cos 78^o} - \frac{\cos 18^o}{\sin 72^o}$    [2003]

$\frac{\cos 75^o}{\sin 15^o} + \frac{\sin 12^o}{\cos 78^o} - \frac{\cos 18^o}{\sin 72^o}$

$=$ $\frac{\cos 75^o}{\sin (90^o - 75^o)} + \frac{\sin 12^o}{\cos (90^o - 12^o)} - \frac{\cos 18^o}{\sin (90^o - 18^o)}$

$=$ $\frac{\cos 75^o}{\cos 75^o} + \frac{\sin 12^o}{\sin 12^o} - \frac{\cos 18^o}{\cos 18^o}$

$= 1+ 1 - 1 = 1$

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Question 22: Prove that:  $\frac{\sin A}{1 + \cos A} + \frac{1 +\cos A}{\sin A}$ $= 2 \ cosec \ A$    [2009]

LHS $=$ $\frac{\sin A}{1 + \cos A} + \frac{1 +\cos A}{\sin A}$

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

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

$=$ $\frac{2(1+\cos A)}{(1+ \cos A)\sin A}$

$=$ $\frac{2}{\sin A}$

$=$ $2 \ cosec A =$ RHS.

Hence proved.

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Question 23: Prove that: $\frac{\cos A \cot A}{1 - \sin A}$ $= 1 + cosec \ A$    [2006]

LHS $=$ $\frac{\cos A. \cot A}{1 - \sin A}$

$=$ $\frac{\cos A . \frac{\cos A}{\sin A}}{1 - \sin A}$

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

$=$ $\frac{(1-\sin A)(1 + \sin A)}{\sin A(1 - \sin A)}$

$=$ $\frac{1 + \sin A}{\sin A}$

$= cosec \ A + 1 =$ RHS. Hence proved.

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Question 24: Without using trigonometric tables evaluate :

$\frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{cosec^2 10^o - \tan^2 80^o}$    [2010]

Given: $\frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{cosec^2 10^o - \tan^2 80^o}$

$=$ $\frac{\sin 35^o \cos (90^o- 35^o) + \cos 35^o \sin (90^o- 35^o)}{cosec^2 10^o - \tan^2 (90^o - 10^o)}$

$=$ $\frac{\sin^2 35^o+\cos^2 35^o}{1+ \cot^2 10^o - \cot^2 10^o}$

$=$ $\frac{1}{1}$

$= 1$

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Question 25:  Without using trigonometric tables evaluate :

$\frac{\sin \ 35^o \cos \ 55^o + \cos \ 35^o \sin \ 55^o}{cosec^2 \ 10^o - \tan^2 \ 80^o}$       [2010]

Given: $\frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{cosec^2 10^o - \tan^2 80^o}$
$=$ $\frac{\sin 35^o \cos (90^o- 35^o) + \cos 35^o \sin (90^o- 35^o)}{cosec^2 10^o - \tan^2 (90^o - 10^o)}$
$=$ $\frac{\sin^2 35^o+\cos^2 35^o}{1+ \cot^2 10^o - \cot^2 10^o}$
$=$ $\frac{1}{1}$ $= 1$
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