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{\mathrm{cosec} A + 1 }{\mathrm{cosec} A - 1}

Answer:

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

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: \mathrm{cosec}^4 A -  \mathrm{cosec}^2 A = \cot^4 A + \cot^2 A

Answer:

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

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

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

Answer:

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

Answer:

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

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  \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

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

Answer:

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

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  \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}

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}{(\mathrm{cosec} A - 1)^2}

Answer:

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 

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

= \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   

Answer:

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

Answer:

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}

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} = \mathrm{cosec} A - \cot A    [2008]

Answer:

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 

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 -  \mathrm{cosec} A)(1+\tan\ A+\sec\ A)=2 

Answer:

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 

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}} = \mathrm{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} 

= \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]

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

Answer:

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]

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} = \mathrm{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}

= \mathrm{cosec} \theta - 1 = RHS. Hence proved.

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