Problem
If \( x < 0 \), then express:
\[ \cos^{-1}\left(\frac{1 – x^2}{1 + x^2}\right) \] in terms of \( \tan^{-1}x \).
Solution
Let
\[ \tan^{-1}x = \theta \]
Then,
\[ x = \tan \theta \]
Using identity:
\[ \cos 2\theta = \frac{1 – \tan^2\theta}{1 + \tan^2\theta} \]
Substitute \( x = \tan \theta \):
\[ \frac{1 – x^2}{1 + x^2} = \cos 2\theta \]
So,
\[ \cos^{-1}\left(\frac{1 – x^2}{1 + x^2}\right) = \cos^{-1}(\cos 2\theta) \]
Now, since \( x < 0 \Rightarrow \theta = \tan^{-1}x < 0 \), so \( 2\theta < 0 \).
We know that:
\[ \cos^{-1}(\cos y) = -y \quad \text{for } y < 0 \]
Thus,
\[ \cos^{-1}(\cos 2\theta) = -2\theta \]
Substitute back \( \theta = \tan^{-1}x \):
Final Answer:
\[ \boxed{ \cos^{-1}\left(\frac{1 – x^2}{1 + x^2}\right) = -2\tan^{-1}x } \]
Key Concept
For \( x \geq 0 \), the identity is \( 2\tan^{-1}x \), but for \( x < 0 \), sign changes due to principal value range of inverse cosine. :contentReference[oaicite:0]{index=0}