Question

The range of \( x \) for which the equation \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) \) holds true

• A. \( |x|\leq 1 \)
• B. \( \forall x\in R \)
• C. \( x\geq 0 \)
• D. \( |x|\geq 1 \)

Hint:

While using inverse trigonometric identities, always check the principal value range before directly cancelling inverse and trigonometric functions.

Answer 1

The Correct Option is A

Step 1: Use the identity involving tangent.
We know that:
\[ \sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}. \]

Step 2: Put \(\theta=\tan^{-1}x\).
If:
\[ \theta=\tan^{-1}x, \]
then:
\[ \tan\theta=x. \]
Therefore:
\[ \sin(2\tan^{-1}x)=\frac{2x}{1+x^2}. \]

Step 3: Rewrite the given expression.
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin(2\tan^{-1}x)). \]

Step 4: Use principal value range of \(\sin^{-1}x\).
The principal value range of \(\sin^{-1}x\) is:
\[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \]
So,
\[ \sin^{-1}(\sin \theta)=\theta \] only when
\[ \theta \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \]

Step 5: Apply this condition.
Here:
\[ \theta=2\tan^{-1}x. \]
So, we need:
\[ -\frac{\pi}{2}\leq 2\tan^{-1}x \leq \frac{\pi}{2}. \]

Step 6: Divide by 2.
\[ -\frac{\pi}{4}\leq \tan^{-1}x \leq \frac{\pi}{4}. \]
Taking tangent throughout:
\[ -1\leq x \leq 1. \]

Step 7: Write in modulus form.
\[ |x|\leq 1. \]
Final Answer:
\[ \boxed{|x|\leq 1} \]