Question

Solution of the equation

3 sin^–1$\left( \frac{2x}{1 + x^{2}} \right)$– 4 cos^–1$\left( \frac{1 - x^{2}}{1 + x^{2}} \right)$+ 2 tan^–1$\left( \frac{2x}{1 - x^{2}} \right)$= $\frac{\pi}{3}$is-

• A. x = $\sqrt{3}$
• B. x =$\frac{1}{\sqrt{3}}$
• C. x = 1
• D. x = 0

Answer 1

Answer: (B)

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

Answer 2

Given equation is 3.2 tan^–1 x – 4.2 tan^–1 x + 2.2 tan^–1

x = $\frac{\pi}{3}$ ̃ 2 tan^{– 1} x :$\frac{\pi}{3}$ ̃ x = tan $\frac{\pi}{6}$ = $\frac{1}{\sqrt{3}}$