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Question and Answer

\( \frac{tan\theta }{cot(90°-\theta )}+\frac{cot\theta }{tan(90°-\theta )}-2cos\theta\;cosec(90°-\theta )\) is equal to
(A) 2
(B) 1
(C) 0
(D) 3

Answer

(C)
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Solution

\( \therefore\;cot\left(90°-\theta \right)=tan\theta ,tan\left(90°-\theta \right)=cot\theta \) and
\( cosec\left(90°-\theta \right)=sec\theta  \)
\( =\frac{tan\theta }{cot(90°-\theta )}+\frac{cot\theta }{tan(90°-\theta )}-2coscosec (90°-\theta )\)
\( =\frac{tan\theta }{tan\theta }+\frac{cot\theta }{cot\theta }-2cossec\theta  \)
\( =1+1-2cos\frac{1}{cos}=1+1-2=0\)
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