Question

If $\theta ϵ[-\frac{\pi }{2},\frac{\pi }{2}]$ the solution of the equation ${\log }_{\sin \theta }({\cos }^2\theta -{\sin }^2\theta )=2$
is given by

• A. $\theta ={\sin }^{-1}(-\frac{1}{\sqrt{3}})$
• B. $\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. $\theta =n\pi$
• D. $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (B), (D)

The correct options are
B $\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
D $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

By simplification,we get

${\cos }^2\theta -{\sin }^2\theta ={\sin }^2\theta$

Or

${\cos }^2\theta =2{\sin }^2\theta$

$1-{\sin }^2\theta =2{\sin }^2\theta$

$3{\sin }^2\theta =1$

${\sin }^2\theta =\frac{1}{3}$

$\sin \theta =\frac{\pm 1}{\sqrt{3}}$.

since in $\ln \sin \theta$ $<0$ is not possible.

Hence

$\sin \theta =\frac{1}{\sqrt{3}}$

$\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Consider again

${\cos }^2\theta =2{\sin }^2\theta$

Or

${\tan }^2\theta =\frac{1}{2}$
Hence

$\tan \theta =\frac{\pm 1}{\sqrt{2}}$.

$\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$ and

$\theta ={\tan }^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ where $\cos \theta <0$ and $\sin \theta >0$.