Question

The value of \(2\tan^{-1}\left(\frac{1}{3}\right) + \cot^{-1}\left(\frac{3}{4}\right)\) is

• A. $\frac{\pi}{3}$
• B. $\frac{2\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$
• E. $\frac{\pi}{2}$

Hint:

Check $ab=1$ condition quickly to directly conclude $\frac{\pi}{2}$.

Answer 1

Answer: (E)

Concept: Use identities: \[ \cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right), \quad 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

Step 1: Convert $\cot^{-1}$ to $\tan^{-1}$
\[ \cot^{-1}\left(\frac{3}{4}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]

Step 2: Simplify $2\tan^{-1}\left(\frac{1}{3}\right)$
\[ 2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2\cdot \frac{1}{3}}{1 - \frac{1}{9}}\right) = \tan^{-1}\left(\frac{2/3}{8/9}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

Step 3: Add the angles
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{4}{3}\right) \] Using identity: \[ \tan^{-1}a + \tan^{-1}b = \frac{\pi}{2} \quad \text{if } ab=1,\; a,b>0 \] Here: \[ \frac{3}{4} \cdot \frac{4}{3} = 1 \] \[ \Rightarrow \frac{\pi}{2} \] Final Conclusion:
Option (E)