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

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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]

Answer:

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

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

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

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

Answer:

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

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

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

$= 2$

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

Answer:

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

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

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

$= 3 + 2 = 5$

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

Answer:

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

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

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

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

Answer:

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

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

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

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

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

$= 10-5 = 5$

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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]

Answer:

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

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

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

$= 2 + 1 - 3 = 0$

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

Answer:

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

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

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

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

= $\frac{sin \ A}{cos \ A}$ $= tan \ A =$ 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]

Answer:

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]

Answer:

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]

Answer:

$2 sin \ A - 1 = 0$

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

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]

Answer:

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

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

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

$= 3 - 1 = 2$

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

Answer:

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

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

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

$= 3 + 2 = 5$

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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]

Answer:

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

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

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

$= 1+ 1 - 1 = 1$

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

Answer:

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.

$\\$

Question 22: Prove that: $\frac{cos \ A cot \ A}{1 - sin \ A}$ $= 1 + cosec \ A$    [2006]

Answer:

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 23: 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]

Answer:

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

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

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

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

$= 1$

$\\$

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]

Answer:

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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