Prove the following identities:

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: \frac{1+sin \ A }{1- sin \ A} = \frac{cosec \ A + 1 }{cosec \ A - 1}

Answer:

RHS = \frac{cosec \ A + 1 }{cosec \ A - 1}

= \frac{\frac{1}{sin \ A} +1 }{\frac{1}{sin \ A} - 1}

= \frac{1+sin \ A }{1- sin \ A} = LHS. Hence Proved.

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Question 3: \frac{1}{tan \ A + cot A} = cos \ A sin \ A

Answer:

LHS = \frac{1}{tan \ A + cot A}

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

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

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

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Question 4: tan \ A - cot A = \frac{1- 2 \ cos^2 A}{sin \ A . cos \ A}

Answer:

LHS = tan \ A - cot A

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

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

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

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

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Question 5: sin^4 A - cos^4 A = 2 sin^2 A - 1

Answer:

LHS = sin^4 A - cos^4 A

= (sin^2 A - cos^2 A)(sin^2 A + cos^2 A)

= sin^2 A - cos^2 A

= 1- cos^2 A - cos^2 A = 1- 2 \ cos^2 A = RHS. Hence proved.

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Question 6: (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 7: cosec^4 A - cosec^2 A = cot^4 A + cot^2 A

Answer:

LHS = cosec^4 A - cosec^2 A

= \frac{cos^4 A}{sin^4 A} + \frac{cos^2 A}{sin ^2 A}

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

= \frac{cos^2 A}{sin^4 A}

= \frac{1 - sin^2 A}{sin^4 A}

= \frac{1}{sin^4 A} - \frac{1}{sin^2 A}

= cosec^4 A - cosec^2 A = RHS. Hence Proved.

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Question 8: sec \ A(1- sin \ A) (sec \ A + tan \ A) = 1

Answer:

LHS = sec \ A(1- sin \ A) (sec \ A + tan \ A)

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

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

= \frac{cos^2 A}{cos^2 A} = 1 = RHS. Hence Proved.

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Question 9: cosec \ A (1+ cos \ A) (cosec \ A - cot A) = 2

Answer:

LHS = cosec \ A (1+ cos \ A) (cosec \ A - cot A)

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

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

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

=\frac{sin^2 A}{sin^2 A} = 1 = RHS

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Question 10: sec^2 A + cosec^2 A = sec^2  A. cosec^2 A

Answer:

LHS = sec^2 A + cosec^2 A

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

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

= sec^2 A . cosec^2 A = RHS Hence proved.

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Question 11: \frac{(1+tan^2 A) cot A}{cosec^2 A} = tan \ A

Answer:

LHS = \frac{(1+tan^2 A) cot A}{cosec^2 A}

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

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

= tan \ A = RHS. Hence proved.

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Question 12: tan^2 A - sin^2 A= tan^2 A. sin^2 A

Answer:

LHS = tan^2 A - sin^2 A

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

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

= \frac{sin^4 A}{cos^2 A} = tan^2 A. sin^2 A= RHS. Hence proved.

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Question 13: cot^2 A - cos^2 A = cos^2 A .cot^2 A

Answer:

LHS = cot^2 A - cos^2 A

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

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

= \frac{cos^4 A}{sin^2 A} = cot^2 A. cos^2 A= RHS. Hence proved.

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Question 14: (cosec \ A + sin \ A)(cosec \ A - sin \ A) = cot^2 A + cos^ A

Answer:

LHS = (cosec \ A + sin \ A)(cosec \ A - sin \ A)

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

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

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

= cot^2 A + cos^2 A= RHS. Hence Proved.

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Question 15: (sec \ A - cos \ A)(sec \ A + cos \ A)= sin^2 A + tan^2 A

Answer:

LHS = (sec \ A - cos \ A)(sec \ A + cos \ A)

= (\frac{1}{cos \ A} + cos \ A) ( \frac{1}{cos \ A} - cos \ A)

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

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

= tan^2 A + sin^2 A = RHS. Hence Proved.

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Question 16: (cos \ A + sin \ A)^2 + (cos \ A - sin \ A)^2 = 2

Answer:

LHS = (cos \ A + sin \ A)^2 + (cos \ A - sin \ A)^2

= cos^2 A + sin^2 A + 2 \ cos \ A .sin \ A + cos^2 A + sin^2 A - 2 \ cos \ A .sin \ A

= 2(cos^2 A + sin^2 A) = RHS. Hence proved.

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Question 17: (cosec \ A - sin \ A) (sec \ A - cos \ A) (tan \ A + cot A) = 2

Answer:

LHS = (cosec \ A - sin \ A) (sec \ A - cos \ A) (tan \ A + cot A)

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

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

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

= 1 = RHS. Hence proved.

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Question 18: \frac{1}{sec \ A + tan \ A} = sec \ A - tan \ A

Answer:

LHS = \frac{1}{sec \ A + tan \ A}

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

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

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

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

= \frac{1-sin \ A}{cos \ A} = sec \ A - tan \ A = RHS

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Question 19: cosec \ A + cot A = \frac{1}{cosec \ A - cot A}

Answer:

LHS = 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{sin \ A}{1 - cos \ A}

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

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

Therefore LHS = RHS. Hence proved.

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Question 20: \frac{sec \ A - tan \ A}{sec \ A + tan \ A} = 2 sec^2 A - 1-2 sec \ A tan \ A

Answer:

LHS = \frac{sec \ A - tan \ A}{sec \ A + tan \ A}

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

= \frac{1-sin \ A}{1+ sin \ A}

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

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

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

= 2 sec^2 A - 1-2 sec \ A . tan \ A

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Question 21: (sin \ A + cosec \ A)^2 +(cos \ A + sec \ A)^2 = 7 + tan^2 A + cot^2 A

Answer:

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

= sin^2 A + cosec^2 A + 2 + cos^2 A + sec^2 A + 2

= 5 + cosec^2 A + sec^2 A

= 5 + (1 + cot^2 A) + (1 + tan^2 A)

= 7 + tan^2 A + cot^2 A = RHS. Hence proved.

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Question 22: sec^2 A . cosec^2 A = tan^2 A + cot^2 A + 2

Answer:

LHS = sec^2 A . cosec^2 A

= (1 +tan^2 A)(1 + cot^2 A)

= 1 +tan^2 A+ cot^2 A + 1

= tan^2 A + cot^2 A + 2 =  RHS. Hence proved.

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Question 23: \frac{1}{1+cos \ A} + \frac{1}{1- cos \ A} = 2 cosec^2 A

Answer:

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

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

= \frac{2}{sin^2 A} = 2 cosec^2 A = RHS. Hence proved.

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Question 24: \frac{1}{1 - sin \ A} + \frac{1}{1+sin \ A} = 2 sec^2 A

Answer:

LHS = \frac{1}{1 - sin \ A} + \frac{1}{1+sin \ A}

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

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

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