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

RHS $=$ $\frac{ \mathrm{cosec} A + 1 }{\mathrm{cosec} A - 1}$  $=$ $\frac{\frac{1}{\sin A} +1 }{\frac{1}{\sin A} - 1}$  $=$ $\frac{1+\sin A }{1- \sin A}$ $=$ 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: $\mathrm{cosec}^4 A - \mathrm{cosec}^2 A = \cot^4 A + \cot^2 A$

LHS $= \mathrm{cosec}^4 A - \mathrm{cosec}^2 A$

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

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

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

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

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

$= \mathrm{cosec}^4 A - \mathrm{cosec}^2 A =$ 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: $\mathrm{cosec} A (1+ \cos A) (\mathrm{cosec} A - \cot A) = 2$

LHS $= \mathrm{cosec} A (1+ \cos A) (\mathrm{cosec} A - \cot A)$

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

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

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

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

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Question 10: $\sec^2 A + \mathrm{cosec}^2 A = \sec^2 A. \mathrm{cosec}^2 A$

LHS $= \sec^2 A + \mathrm{cosec}^2 A$

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

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

$= \sec^2 A . \mathrm{cosec}^2 A = RHS$ Hence proved.

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

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

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

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

$= \tan A =$ 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: $(\mathrm{cosec} A + \sin A)(\mathrm{cosec} A - \sin A) = \cot^2 A + \cos^ A$

LHS $= (\mathrm{cosec} A + \sin A)(\mathrm{cosec} A - \sin A)$

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

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

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

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

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

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

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

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

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

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

LHS $= (\mathrm{cosec} A - \sin A) (\sec A - \cos A) (\tan A + \cot A)$

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

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

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

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

LHS $= \mathrm{cosec} A + \cot A$

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

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

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

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

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

LHS $= (\sin A + \mathrm{cosec} A)^2 +(\cos A + \sec A)^2$

$= \sin^2 A + \mathrm{cosec}^2 A + 2 + \cos^2 A + \sec^2 A + 2$

$= 5 + \mathrm{cosec}^2 A + \sec^2 A$

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

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

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Question 22: $\sec^2 A . \mathrm{cosec}^2 A = \tan^2 A + \cot^2 A + 2$

LHS $= \sec^2 A . \mathrm{cosec}^2 A$

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

$= 1 +\tan^2 A+ \cot^2 A + 1$

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

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

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

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

$=$ $\frac{2}{\sin^2 A}$ $= 2 \mathrm{cosec}^2 A =$ 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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Question 25:    $\frac{\mathrm{cosec} A}{\mathrm{cosec} A - 1}$ $+$ $\frac{\mathrm{cosec} A}{\mathrm{cosec} A + 1}$ $= 2 \sec^2 A$

LHS $=$ $\frac{\mathrm{cosec} A}{\mathrm{cosec} A - 1}$ $+$ $\frac{\mathrm{cosec} A}{\mathrm{cosec} A + 1}$

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

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

$=$ $\frac{2}{\cos^2 A}$

$=$ $2 \sec^2 A =$ RHS. Hence proved.

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Question 26:    $\frac{\sec A}{\sec A - 1}$ $+$ $\frac{\sec A}{\sec A + 1}$ $= 2 \mathrm{cosec}^2 A$

LHS $=$ $\frac{\sec A}{\sec A - 1}$ $+$ $\frac{\sec A}{\sec A + 1}$

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

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

$=$ $\frac{2}{\sin^2 A}$

$=$ $2 \mathrm{cosec}^2 A =$ RHS. Hence proved.

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Question 27:   $\frac{1+ \cos A}{1- \cos A}$ $=$ $\frac{\tan^2 A}{(\sec A - 1)^2}$

RHS $=$ $\frac{\tan^2 A}{(\sec A - 1)^2}$

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

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

$=$ $\frac{1+ \cos A}{1- \cos A} =$ LHS. Hence proved.

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Question 28:   $\frac{1- \sin A}{1 + \sin A}$ $=$ $\frac{\cot^2 A}{(\mathrm{cosec} A - 1)^2}$

RHS $=$ $\frac{\cot^2 A}{(\mathrm{cosec} A - 1)^2}$

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

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

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

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Question 29:   $\frac{1 + \sin A}{\cos A}$ $+$ $\frac{\cos A}{1 + \sin A}$ $= 2 \sec A$

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

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

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

$=$ $\frac{2}{\cos A}$

$= 2 \sec A =$ RHS. Hence proved.

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Question 30:   $\frac{1- \sin A}{1 + \sin A}$ $= (\sec A - \tan A)^2$

RHS $=$ $(\sec A - \tan A)^2$

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

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

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

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Question 31:   $\frac{1- \cos A}{1 + \cos A}$ $= (\cot A - \mathrm{cosec} A)^2$

RHS $=$ $(\cot A - \mathrm{cosec} A)^2$

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

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

$=$ $\frac{1- \cos A}{1+ \cos A}$ $=$ LHS. Hence proved.

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Question 32:   $\frac{\mathrm{cosec} A - 1}{\mathrm{cosec} A + 1}$ $= ($ $\frac{\cos A}{1 + \sin A}$ $)^2$

LHS $=$ $\frac{\mathrm{cosec} A - 1}{\mathrm{cosec} A + 1}$

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

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

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

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

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Question 33:   $\tan^2 A - \tan^2 B$ $=$ $\frac{\sin^2 A - \sin^2 B}{\cos^2 A. \cos^2 B}$

LHS $=$ $\tan^2 A - \tan^2 B$

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

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

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

$=$ $\frac{\sin^2 A - \sin^2 B}{\cos^2 A. \cos^2 B} =$ RHS. Hence proved.

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Question 34:   $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$ $= \tan A$

LHS $=$ $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$

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

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

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

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

$=$ $\tan A =$ RHS. Hence proved.

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Question 35:   $\frac{\sin A}{1 + \cos A}$ $= \mathrm{cosec} A - \cot A$   [2008]

RHS $=$ $\mathrm{cosec} A - \cot A$

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

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

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

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

$=$ $\frac{\sin A}{1 + \cos A}$ $=$ LHS. Hence proved.

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Question 36:   $\frac{\cos A}{1 - \sin A}$ $= \sec A + \tan A$

RHS $=$ $\sec A + \tan A$

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

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

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

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

$=$ $\frac{\cos A}{1 - \sin A}$ $=$ LHS. Hence proved.

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Question 37:   $\frac{\sin A. \tan A}{1- \cos A}$ $= 1+ \sec A$

LHS $=$ $\frac{\sin A. \tan A}{1- \cos A}$

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

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

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

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

$=$ $1 + \sec A$

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Question 38:   $(1+\cot\ A - \mathrm{cosec} A)(1+\tan\ A+\sec\ A)=2$

LHS $=$ $(1+\cot\ A - \mathrm{cosec} A)(1+\tan\ A+\sec\ A)$

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

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

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

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

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

$= 2 =$ RHS. Hence proved.

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Question 39:   $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$ $= \sec A + \tan A$

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

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

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

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

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

$=$ $\sec A + \tan A =$ RHS. Hence proved.

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Question 40:   $\sqrt{\frac{1 - \cos A}{1 + \cos A}}$ $= \mathrm{cosec} A - \cot A$    [2000]

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

$=$ $\sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}}$

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

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

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

$=$ $\mathrm{cosec} A - \cot A =$ RHS. Hence proved.

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Question 41:   $\sqrt{\frac{1 - \cos A}{1 + \cos A}}$ $=$ $\frac{\sin A}{1 + \cos A}$     [2013]

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 42:   $\sqrt{\frac{1 - \sin A}{1 + \sin A}}$ $=$ $\frac{\cos A}{1 + \sin A}$

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

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

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

$=$ $\frac{\cos A}{1 + \sin A}$ = RHS. Hence proved.

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Question 43:   $1 -$ $\frac{\cos^2 A}{1 + \sin A}$ $= \sin A$    [2001]

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 44:   $\frac{1}{\sin A + \cos A}$ $+$ $\frac{1}{\sin A + \cos A}$ $=$ $\frac{2 \sin A}{1 - 2 \cos^2 A}$    [2002]

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

LHS $=$ $\frac{\sin A + \cos A}{\sin A - \cos A}$ $+$ $\frac{\sin A - \cos A}{\sin A + \cos A}$

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

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

$=$ $\frac{2}{1 - 2. \cos^2 A}$ = RHS. Hence proved.

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Question 46:   $\frac{\cot A + \mathrm{cosec} A - 1}{\cot A - \mathrm{cosec} A + 1}$ $=$ $\frac{1 + \cos A}{\sin A}$

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

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

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

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

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

$=$ $\frac{1 + \cos A}{\sin A}$ = RHS. Hence proved.

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Question 47:   $\frac{\sin \theta . \tan \theta}{1 - \cos \theta}$ $= 1 + \sec \theta$    [2006]

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 48:   $\frac{\cos \theta . \cot \theta}{1 + \sin \theta}$ $= \mathrm{cosec} \theta - 1$

LHS $=$ $\frac{\cos \theta . \cot \theta}{1 + \sin \theta}$
$=$ $\frac{\cos^2 \theta}{\sin \theta (1 + \sin \theta)}$
$=$ $\frac{(1-\sin \theta)(1 + \sin \theta)}{\sin \theta (1 + \sin \theta)}$
$=$ $\frac{1 - \sin \theta}{\sin \theta}$
$=$ $\mathrm{cosec} \theta - 1$ = RHS. Hence proved.
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