Question

The value of $ \cos \left( \frac{3\pi }{2}+x \right)\,\cos \,(2\pi +x)\left\{ \cot \left( \frac{3\pi }{2}-x \right)+\cot \,(2\pi +x) \right\} $ is

• A. $ 0 $
• B. $ 1 $
• C. $ cos\text{ }x $
• D. $ sin\text{ }x $

Answer 1

Answer: (B)

$ \cos \left( \frac{3\pi }{2}+x \right)\cos (2\pi +x)\, $
$ \left\{ \cot \left( \frac{3\pi }{2}-x \right)\,+\cot \,(2\pi +x) \right\} $ $ =\sin x.\cos x\,(\tan \,x+\,\cot \,x\} $
$ =\sin x.\cos x\left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x.\cos x} \right) $
$ =1 $
$ (\because \,\,\,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1) $