Question

The number of solutions of the equation, $sin^{-1}\, x = 2 tan^{-1}\,x$ (in principal values) is:

• A. $1$
• B. $4$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$
\begin{array}{l}
\sin ^{-1} x=2 \tan ^{-1} x \\
\sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \\
x=\frac{2 x}{1+x^{2}} \\
x\left(1+x^{2}\right)=2 x \\
x+x^{3}=2 x \\
x^{3}-x=0
\end{array}
$
Hence $0,-1,1$ are
solution of equation
i.e. $x\left(x^{2}-1^{2}\right)=0$
$
x(x+1)(x-1)=0
$
$
x=0, x+1=0, x-1=0
$
$
x=-1, x=1
$
$\therefore$ It has 3 solutions.