Question

For $\theta$ being acute, if $\sin 3\theta =1$ find the value of $\cot \theta -\tan 2\theta$.

Answer 1

Given,

$\sin 3\theta =1$

or, $3\theta =\frac{\pi }{2}$ [ Since $\theta$ being acute]

or, $\theta =\frac{\pi }{6}$.

Now,

$\cot \theta -\tan 2\theta$

$=\cot \frac{\pi }{6}-\tan \frac{\pi }{3}$

$=\sqrt{3}-\sqrt{3}$

$=0$.