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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$= 2 =$ 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:   $\sqrt{\frac{1 - cos \ A}{1 + cos \ A}}$ $= cosec \ A - cot \ A$    [2000]

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

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

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

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

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

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

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

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

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

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

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

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

LHS $=$ $\frac{cos \ \theta . cot \ \theta}{1 + sin \ \theta}$

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

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

$=$ $\frac{1 - sin \ \theta}{sin \ \theta}$

$=$ $cosec \ \theta - 1$ = RHS. Hence proved.

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