Question

If value of $x$ which satisfy equation ${\sin }^{-1}\frac{2x}{1+x^2}={\tan }^{-1}\frac{2x}{1-x^2}$, is $-a<x<b$.
Find the value of $a+b$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We know, $-1\le \frac{2x}{1+x^2}\le 1$

Now, $-(1+x^2)\le 2x$

$1+x^2+2x\ge 0$

$(1+x{)}^2\ge 0$

$1+x\ge 0$

$x\ge -1$ ...(i)

And $\frac{2x}{1+x^2}\le 1$

$0\le 1+x^2-2x$

$0\le (1-x{)}^2$

$0\le 1-x$

$x\le 1$

Hence, $-1\le x\le 1$.

Hence, $a=-1$ and $b=1$

$\therefore a+b=0$