Question

$\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-ta{n}^{-1}\left(\right(1/3\left)tan\theta \right)$ if $tan\theta$ is equal to

• A. -2
• B. 1
• C. 2/3
• D. 2

Answer 1

Answer: (A), (B)

The correct options are
A 1
B -2

$\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-ta{n}^{-1}\left(\frac{1}{3}tan\theta \right)\Rightarrow \theta =ta{n}^{-1}\left(\frac{2ta{n}^2\theta -\frac{1}{3}tan\theta }{1+\left(2ta{n}^2\theta \right)\left(\frac{1}{3}tan\theta \right)}\right)\Rightarrow tan\theta =\frac{2ta{n}^2\theta -\frac{1}{3}tan\theta }{1+\left(\frac{2}{3}ta{n}^3\theta \right)}\Rightarrow tan\theta (\frac{2tan\theta -\frac{1}{3}}{1+\left(\frac{2}{3}ta{n}^3\theta \right)}-1)=0$
$\Rightarrow tan\theta =0$ or $\frac{2tan\theta -\frac{1}{3}}{1+\left(\frac{2}{3}ta{n}^3\theta \right)}-1=0$
$\frac{2tan\theta -\frac{1}{3}}{1+\left(\frac{2}{3}ta{n}^3\theta \right)}=1\Rightarrow ta{n}^3\theta -3tan\theta +2=0\Rightarrow {(tan\theta -1)}^2(tan\theta +2)=0$
$\Rightarrow tan\theta =1$ or $tan\theta =-2$