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.

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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

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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]

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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