Inverse Trigonometric Functions Ans 2

Q2 - 25 January - Shift 1

If the sum of all the solutions of $\tan ^{-1}(\frac{2 x}{1-x^{2}})+\cot ^{-1}(\frac{1-x^{2}}{2 x})=\frac{\pi}{3}$, $-1<x<1, x \neq 0$, is $\alpha-\frac{4}{\sqrt{3}}$, then $\alpha$ is equal to _______

Show Answer

Answer: 2

Solution:

Formula: Identity of double angle of inverse tangent function : $\tan ^{-1} ( \frac{2x}{1-x^2})$

Case I : $x>0$

$\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}$

$x=2-\sqrt{3}$

Case II : $x<0$

$\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}+\pi=\frac{\pi}{3}$

$x=\frac{-1}{\sqrt{3}}$

$ \Rightarrow \alpha=2$

$\therefore \alpha$ is equal to 2.