Question

If $sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+cos^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right) = tan^{-1} \left(\frac{2x}{1-x^{2}}\right)$, where $a, x \in ]0, 1[$, then the value of $x$ is

• A. $0$
• B. $\frac{a}{2}$
• C. $a$
• D. $\frac{2a}{1+a^{2}}$

Answer 1

Answer: (D)

$sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+cos^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)
= tan^{-1} \left(\frac{2x}{1-x^{2}}\right)$
$\Rightarrow 2tan^{-1}a + 2tan^{-1}a = 2tan^{-1}x$
$\Rightarrow 4tan^{-1}\left(a\right) = 2tan^{-1}x$
$\Rightarrow 2tan^{-1}a = tan^{-1}x$
$\Rightarrow tan^{-1} \left(\frac{2a}{1+a^{2}}\right)= tan^{-1}x$
$\Rightarrow x = \frac{2a}{1+a^{2}}$