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)

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