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

Answer:

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

Answer:

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}

Answer:

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}

Answer:

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 

Answer:

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

Answer:

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   

Answer:

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

Answer:

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}

Answer:

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

Answer:

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]

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

Answer:

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 

Answer:

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 

Answer:

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 

Answer:

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]

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 41:   \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 42:   \sqrt{\frac{1 - sin \ A}{1 + sin \ A}} = \frac{cos \ A}{1 + sin \ A}

Answer:

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]

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 44:   \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 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}

Answer:

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}

Answer:

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]

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

Answer:

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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