Question

If \(3sin^{-1}(\frac{2x}{1+x^2})-4cos^{-1}(\frac{1-x^2}{1+x^2})+2tan^{-1}(\frac{2x}{1-x^2}) = \frac{π}{3}\) then \(x = ?\)

• A. \(\sqrt{3}\)
• B. \(1\)
• C. \(\frac{1}{\sqrt{3}}\)
• D. \(-1\)

Hint:

Convert each inverse function to 2 tan inverse x, valid for x between 0 and 1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Three standard identities hold for \(0\le x\le1\):
\[ \sin^{-1}\frac{2x}{1+x^2}=2\tan^{-1}x,\quad \cos^{-1}\frac{1-x^2}{1+x^2}=2\tan^{-1}x,\quad \tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x \]

Step 2: Substitute
\[ 3(2\tan^{-1}x)-4(2\tan^{-1}x)+2(2\tan^{-1}x)=\frac\pi3 \]

Step 3: Combine
\[ (6-8+4)\tan^{-1}x=2\tan^{-1}x=\frac\pi3 \]
\[ \tan^{-1}x=\frac\pi6\Rightarrow x=\tan\frac\pi6=\frac1{\sqrt3} \]

Step 4: Check
\(x=1/\sqrt3\) lies in the allowed range \(0\le x<1\). The value \(x=\sqrt3\) lies outside the range for these identities, and \(x=-1\) gives a negative left-hand side. Answer: (C).

Final Answer:
The equation reduces to \(2\tan^{-1}x=\pi/3\), so \(x=1/\sqrt3\), option (C). \[ \boxed{\frac{1}{\sqrt{3}}} \]