If -1 < x < 0, then write the value of sin^-1(2x/(1+x^2)) + cos^-1((1-x^2)/(1+x^2)).

If −1 < x < 0, find sin⁻¹(2x/(1+x²)) + cos⁻¹((1−x²)/(1+x²))

Question

If \( -1 < x < 0 \), find the value of:

\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]

Let

\[ x = \tan \theta \]

Then, since \( -1 < x < 0 \Rightarrow \theta \in \left(-\frac{\pi}{4}, 0\right) \)

Using identities:

\[ \sin 2\theta = \frac{2x}{1+x^2} \quad \text{and} \quad \cos 2\theta = \frac{1-x^2}{1+x^2} \]

So the expression becomes:

\[ \sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\theta) \]

Now, since \( \theta \in (-\frac{\pi}{4}, 0) \Rightarrow 2\theta \in (-\frac{\pi}{2}, 0) \)

Using principal values:

\[ \sin^{-1}(\sin 2\theta) = 2\theta \quad (\text{since } -\tfrac{\pi}{2} < 2\theta < \tfrac{\pi}{2}) \]

\[ \cos^{-1}(\cos 2\theta) = -2\theta \quad (\text{since } 2\theta < 0) \]

Therefore,

\[ \sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\theta) = 2\theta + (-2\theta) \]

\[ = 0 \]

Final Answer:

\[ \boxed{0} \]

Careful use of principal value ranges converts the expression into a simple cancellation.

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