Prove the following identities:

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

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

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

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: $sin^4 A - cos^4 A = 2 sin^2 A - 1$

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]

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: $cosec^4 A - cosec^2 A = cot^4 A + cot^2 A$

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

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

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: $sec^2 A + cosec^2 A = sec^2 A. cosec^2 A$

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

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: $tan^2 A - sin^2 A= tan^2 A. sin^2 A$

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: $cot^2 A - cos^2 A = cos^2 A .cot^2 A$

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

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

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

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

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

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

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

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

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: $sec^2 A . cosec^2 A = tan^2 A + cot^2 A + 2$

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

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

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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