If 0 ˂ A ˂ 90°, then the value of \frac{1}{2} \cot \A [ {1+(\sec A\-\tan A)^{2}}/{\cos {ec} A(\sec A-\tan A)} ] is

SSC CPO SI and ASI 2016 Paper

Question : 115 of 200

Marks: +1, -0

If 0 ˂ A ˂ 90°, then the value of

\ {1}/{2} \cot \A\ [ {1+(\sec A\-\tan A)^{2}}/{\cos {ec} A(\sec A-\tan A)}\ ]

Solution:

(3)

\ {1}/{2} \cot A\ [\ {1+(\sec A-\tan A)^{2}}/{\ cosec\ A(\sec A-\tan A)}\ ]

= {1}/{2} \cot A\ [\ {1+\sec ^{2} A+\tan ^{2} A-2 \sec A. \tan A}/{\cosec A(\sec A-\tan A)}\ ]

= {1}/{2} \cot A\ [\ {2 \sec ^{2} A-2 \secA .\tan A}/{\cosec A(\sec A-\tan A)}\ ]

= \ {1}/{2} \cot A\ [\ {2 \sec A(\sec A-\tan A)}/{\cosec A(\sec A-\tan A)}\ ]

=\ {1}/{2} \cot A\ (\ {2 \sec A}/{\cosec A} )

=\cot A \tan A= {\cos A}/{\sin A} \ × {\sin A}/{\cos A}=1

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