Question

Evaluate
$\sin ({180}^{\circ }-x).\cot ({270}^{\circ }+x).\tan ({270}^{\circ }-x).\sec ({180}^{\circ }+x)$

• A. $\sec x$
• B. $\cot x$
• C. $\sin x$
• D. $\tan x$

Answer 1

The correct option is D $\tan x$

We have, $\sin ({180}^{\circ }-x).\cot ({270}^{\circ }+x).\tan ({270}^{\circ }-x).\sec ({180}^{\circ }+x)$
$(\because \sin ({180}^{\circ }-x)=\sin x,\cot ({270}^{\circ }+x)=(-\tan x\left)\text{and}\tan \right({270}^{\circ }-x)=\cot x,\sec ({180}^{\circ }+x)=-\sec x)$
$\Rightarrow \sin x.(-\tan x).\cot x.(-\sec x)$
$(\because \tan x.\cot x=1)$
$\Rightarrow \sin x.\sec x=\tan x$