Question

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

• A. $-2$
• B. $1$
• C. $\frac{2}{3}$
• D. 2

Answer 1

Answer: (A), (B)

$1$

$\theta =ta{n}^{-1}\left(2ta{n}^2\theta \right)-ta{n}^{-1}\left(\left(\frac{1}{3}\right)tan\theta \right)\Rightarrow tan\theta =\frac{2ta{n}^2\theta -\left(\frac{1}{3}\right)tan\theta }{1+\left(\frac{2}{3}\right)ta{n}^3\theta }\Rightarrow tan\theta [\frac{2tan\theta -\left(\frac{1}{3}\right)}{1+\left(\frac{2}{3}\right)ta{n}^3\theta }-1]=0$
which is true if $tan\theta =0$ or
$\frac{2tan\theta -\left(\frac{1}{3}\right)}{1+\left(\frac{2}{3}\right)ta{n}^3\theta }=1\Rightarrow ta{n}^3\theta -3tan\theta +2=0$
$\Rightarrow (tan\theta -1{)}^2(tan\theta +2)=0$ which holds if
$tan\theta =1ortan\theta =-2$