Question

$\frac{\sin 18°}{\cos 72°}\times \frac{\tan 150°}{cot60°}$ is equal to

• A.
  -$\sqrt{3}$
• B.
  -$1$
• C.
  $\frac{1}{2}$
• D.
  $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B

-$1$
we can see that the angles $18°\mathrm{and}72°$ are complimentary.
$\Rightarrow \sin 18°=\sin (90°-72°)\Rightarrow \sin 18°=\cos 72°$
Using the relation $\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta$.
Thus, our relation becomes
$\frac{\sin 18°}{\cos 72°}\times \frac{\tan 150°}{\cot 60°}=\frac{\cos 72°}{\cos 72°}\times \frac{\tan 150°}{\cot 60°}\Rightarrow \frac{\sin 18°}{\cos 72°}\times \frac{\tan 150°}{\cot 60°}=\frac{\tan 150°}{\cot 60°}$
Also $\tan 150°=\tan (90°+60°)\Rightarrow \tan 150°=-\cot 60°$
Since, $\tan (90°+\theta )=-\cot \theta$
Thus, our expression becomes
$\frac{\sin 18°}{\cos 72°}\times \frac{\tan 150°}{\cot 60°}=-\frac{\cot 60°}{\cot 60°}=-1$
Thus, option b. is the correct option.