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 

= 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]

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 23: 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 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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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]

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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