Question

The value of \theta for which $\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}si{n}^{-1}\left(\frac{3sin2\theta }{5+4sin2\theta }\right)$

• A.
  
• B.
  
• C.
  
• D. None of these

Answer 1

The correct option is B

$\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}si{n}^{-1}\left(\frac{6tan\theta }{9+ta{n}^2\theta }\right)$
$=ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}si{n}^{-1}\left\{\frac{2\left(\frac{1}{3}tan\theta \right)}{1+{\left(\frac{1}{3}tan\theta \right)}^2}\right\}$
$=ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{1}{2}si{n}^{-1}\left\{\frac{2\times \frac{1}{3}tan\theta }{1+{\left(\frac{1}{3}tan\theta \right)}^2}\right\}$
$=ta{n}^{-1}\left(2ta{n}^2\theta \right)-\frac{2}{2}ta{n}^{-1}\left(\frac{1}{3}tan\theta \right)$
$ta{n}^{-1}\left(\frac{2ta{n}^2\theta -\frac{1}{3}tan\theta }{1+\frac{2}{3}ta{n}^3\theta }\right)$
$\Rightarrow tan\theta =\frac{6ta{n}^2\theta -tan\theta }{3+2ta{n}^3\theta }$
$\Rightarrow 2ta{n}^4\theta -6ta{n}^2\theta +4tan\theta =0$
$\Rightarrow 2tan(ta{n}^3\theta -3tan\theta +2)=0$
$\Rightarrow 2tan(tan\theta -1{)}^2(tan\theta +2)=0$
$\Rightarrow tan\theta =0,-2,1$
$\Rightarrow \theta =n\pi ,n\pi +\frac{\pi }{4},n\pi +ta{n}^{-1}(-2),n\pi +\frac{\pi }{4}$