Question

If $f(\theta)=\tan \left(\sin ^{-1} \sqrt{\frac{2}{3+\cos 2 \theta}}\right)$, then

• A. $f ^{\prime}\left(\frac{\pi}{4}\right)=\sqrt{2}$.
• B. $f \left(\frac{\pi}{4}\right)=\sqrt{2}$.
• C. $\frac{ d ( f (\theta))}{ d (\cos \theta)}$ at $\theta=\frac{\pi}{4}$ is $-\sqrt{2}$.
• D. $\frac{ d ( f (\theta))}{ d (\cos \theta)}$ at $\theta=\frac{\pi}{4}$ is -2 .

Answer 1

Answer: (A), (B), (D)

$f(\theta)=\tan \left(\sin ^{-1} \sqrt{\frac{2}{3+\cos 2 \theta}}\right)$
$f (\theta)=\sqrt{\frac{2}{1+\cos 2 \theta}}=\sec \theta \Rightarrow f \left(\frac{\pi}{4}\right)=\sqrt{2} . $
$f ^{\prime}(\theta)=\sec \theta \tan \theta$
$f ^{\prime}\left(\frac{\pi}{4}\right)=\sqrt{2}$
$\left.\frac{d(f(\theta))}{d(\cos \theta)}\right|_{\theta=\frac{\pi}{4}}=\left(\frac{-1}{\cos ^2 \theta}\right)_{\theta=\frac{\pi}{4}}=-2$