Inverse Trigonometric Functions Question 2
Question 2 - 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.
