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

Answer

(B)
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Solution

\( \therefore\;sin\left(90°-\theta \right)=cos\theta  \)
\( cos\left(90°-\theta \right)=sin\theta \)
So, \( \frac{sin\theta .cos\theta\;cos(90°-\theta )}{sin(90°-\theta )}+\frac{sin\theta\;cos\theta .sin(90°-\theta )}{cos(90°-\theta )}\)
\( \frac{sin\theta\;cos\theta\;sin\theta }{cos\theta }+\frac{sin\theta\;cos\theta .cos\theta }{sin\theta }\)
\( si{n}^{2}\theta +co{s}^{2}\theta =1        \) \( (\therefore\;si{n}^{2}\theta +co{s}^{2}\theta =1)\)
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