Question

\(2\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\)

• A. \(\tan ^{-1}(18)\)
• B. \(-\tan ^{-1}(18)\)
• C. \(\pi +\tan ^{-1}(18)\)
• D. \(\pi -\tan ^{-1}(18)\)

Hint:

Convert the inverse sine term to an inverse tangent, combine the terms with the addition formula and watch the product of tangents.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Use \(\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4}\) and \(2\tan^{-1}u=\tan^{-1}\frac{2u}{1-u^{2}}\) when \(|u|<1\).

Step 2: Combine the first term
\(2\sin^{-1}\frac{3}{5}=2\tan^{-1}\frac{3}{4}=\tan^{-1}\frac{24}{7}\).
Also \(\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{12/35}{1-1/35}=\tan^{-1}\frac{6}{17}\).

Step 3: Add the two results
The product \(\frac{24}{7}\cdot\frac{6}{17}=\frac{144}{119}>1\), so the sum is \(\pi+\tan^{-1}\frac{\frac{24}{7}+\frac{6}{17}}{1-\frac{144}{119}}=\pi+\tan^{-1}(-18)\).
Thus the value is \(\pi-\tan^{-1}18\).

Final Answer:
The expression equals \(\pi-\tan^{-1}(18)\). \[ \boxed{\text{(D) }\pi -\tan ^{-1}(18)} \]