Inverse Trigonometric Functions Question 80
Question: $ \tan { \frac{1}{2}{{\sin }^{-1}}\frac{2x}{1+x^{2}}+\frac{1}{2}{{\cos }^{-1}}\frac{1-y^{2}}{1+y^{2}} }= $
Options:
A) $ \frac{x-y}{1+xy} $
B) $ \frac{x+y}{1-xy} $
C) $ \frac{x-y}{x+y} $
D) $ \frac{1-xy}{1+xy} $
Show Answer
Answer:
Correct Answer: B
Put $ x=\tan \theta $ and $ y=\tan \phi $
Let $ x = tan\theta \Rightarrow \theta=tan^{−1}x $
$ \therefore sin ^{−1} \frac{2x}{1+x^{2}}=sin^{−1} \frac ({2tan\theta}{1+tan2\theta})={sin^{−1}}(sin2\theta)=2\theta=2tan^{−1}x $
Let $y=tanϕ \Rightarrow ϕ=tan−1y $
$ \therefore cos^{−1} \frac{1−y^{2}}{1+y^2}=cos^{−1} (\frac{{{1−tan^{2}ϕ}}}{{{1+tan^{2}ϕ}}})=cos^{−1}(cos2ϕ)=2ϕ=2tan^{−1}y$
$ \therefore tan\frac{1}{2}[sin^{−1}\frac{2x}{1+x^{2}}+cos^{−1} \frac{1−y^{2}}{1+y^{2}}]$
$=tan\frac{1}{2}[2tan^{−1}x+2tan^{−1}y]$
$=tan[tan^{−1}x+tan^{−1}y]$
$=tan[tan^{−1}(\frac{x+y}{1−xy})]$
$=\frac{x+y}{1−xy}$
BETA
