Question

If $y=ta{n}^{-1}\left(\frac{1-cos3x}{sin3x}\right)$, then $\frac{dy}{dx}=$ .......

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

Answer 1

Answer: (C)

We have,
$y={\tan }^{-1}\left(\frac{1-\cos 3x}{\sin 3x}\right)$
$\Rightarrow y={\tan }^{-1}\left(\frac{2{\sin }^2\left(\frac{3x}{2}\right)}{2\sin \frac{3x}{2}\cos \frac{3x}{2}}\right)$
$\Rightarrow y={\tan }^{-1}\left(\frac{\sin \left(\frac{3x}{2}\right)}{\cos \left(\frac{3x}{2}\right)}\right)$
$\Rightarrow y={\tan }^{-1}\left(\tan \frac{3x}{2}\right)$
$\Rightarrow y=\frac{3x}{2}$
$\therefore \frac{dy}{dx}=\frac{3}{2}$