Question
Evaluate:
\[ 2\tan^{-1}\left\{\csc(\tan^{-1}x) – \tan(\cot^{-1}x)\right\} \]
Solution
Let
\[ \tan^{-1}x = \theta \Rightarrow \tan\theta = x \]
Then,
\[ \sin\theta = \frac{x}{\sqrt{1+x^2}}, \quad \csc\theta = \frac{\sqrt{1+x^2}}{x} \]
So,
\[ \csc(\tan^{-1}x) = \frac{\sqrt{1+x^2}}{x} \]
Now let
\[ \cot^{-1}x = \phi \Rightarrow \cot\phi = x \Rightarrow \tan\phi = \frac{1}{x} \]
So,
\[ \tan(\cot^{-1}x) = \frac{1}{x} \]
Thus expression becomes:
\[ 2\tan^{-1}\left(\frac{\sqrt{1+x^2}}{x} – \frac{1}{x}\right) = 2\tan^{-1}\left(\frac{\sqrt{1+x^2} – 1}{x}\right) \]
Now use identity:
\[ \tan^{-1}\left(\frac{\sqrt{1+x^2} – 1}{x}\right) = \frac{1}{2}\tan^{-1}x \]
Therefore,
\[ 2 \cdot \frac{1}{2}\tan^{-1}x = \tan^{-1}x \]
Final Answer:
\[ \boxed{\tan^{-1}x} \]
Key Concept
Use substitution and standard inverse trigonometric identities for simplification.