Question

$a=ta{n}^{-1}\left(3ta{n}^3\theta \right)$, $b=\frac{1}{2}co{s}^{-1}\left(\frac{5cos2\theta +4}{4+5cos2\theta }\right)$

$\theta =a-b$ if

• A. $tan\theta =0$
• B. $ta{n}^2\theta =\frac{9+\sqrt{33}}{6}$
• C. $ta{n}^2\theta =\frac{9-\sqrt{33}}{6}$
• D. $tan\theta =1$

Answer 1

Answer: (B), (C)

The correct options are
B $ta{n}^2\theta =\frac{9+\sqrt{33}}{6}$
C $ta{n}^2\theta =\frac{9-\sqrt{33}}{6}$

The value of $\frac{5\cos 2\theta +4}{5+4\cos 2\theta }=\frac{5({\cos }^2\theta -{\sin }^2\theta )+4({\cos }^2\theta +{\sin }^2\theta )}{5+4(2{\cos }^2\theta -1)}$

$=\frac{9{\cos }^2\theta -{\sin }^2\theta }{1+8{\cos }^2\theta }=\frac{9-{\tan }^2\theta }{9+{\tan }^2\theta }$

$=\frac{1-{\tan }^2\varphi }{1+{\tan }^2\varphi }$ where $\tan \varphi =\left(1/3\right)\tan \theta$

$=\cos 2\varphi$

$\Rightarrow b=\frac{1}{2}{\cos }^{-1}\left(\cos 2\varphi \right)=\varphi$

$\theta =a-b={\tan }^{-1}\left(3{\tan }^3\theta \right)-\varphi$

$={\tan }^{-1}\left(3{\tan }^3\theta \right)-{\tan }^{-1}\left[\right(1/3\left)\tan \theta \right]$

$\Rightarrow \tan \theta =\frac{3{\tan }^3\theta -\left(1/3\right)\tan \theta }{1+{\tan }^4\theta }$

$\Rightarrow \tan \theta [9{\tan }^2\theta -1-3-3{\tan }^4\theta ]=0$

$\Rightarrow \tan \theta =0$ or $3{\tan }^4\theta -9{\tan }^2\theta +4=0$

${\tan }^2\theta =\frac{9\pm \sqrt{33}}{6}$