Question

Considering only the principal values of the inverse trigonometric function, the value of $\tan \left( \cos^{-1} \frac{1}{5\sqrt{2}} - \sin^{-1} \frac{4}{\sqrt{17}} \right)$ is

• A. $\frac{3}{34}$
• B. $\frac{1}{34}$
• C. $\frac{3}{29}$
• D. $\frac{1}{29}$

Hint:

Convert all inverse ratios to $\tan^{-1}$ to use the simple $\tan(A-B)$ identity.

Answer 1

The Correct Option is D

Step 1: Conversion
Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}} \implies \cos \alpha = \frac{1}{5\sqrt{2}}$.
Then $\tan \alpha = \sqrt{\sec^2 \alpha - 1} = \sqrt{(5\sqrt{2})^2 - 1} = \sqrt{49} = 7$.
Let $\beta = \sin^{-1} \frac{4}{\sqrt{17}} \implies \sin \beta = \frac{4}{\sqrt{17}}$.
Then $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{\sqrt{1 - (4/\sqrt{17})^2}} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.
Step 2: Formula
We need $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Step 3: Calculation
$\tan(\alpha - \beta) = \frac{7 - 4}{1 + (7 \times 4)} = \frac{3}{1 + 28} = \frac{3}{29}$.
Wait, let's re-verify the values. If $\tan \alpha = 7$ and $\tan \beta = 4$, then $\frac{3}{29}$.
Checking the options: (C) is $\frac{3}{29}$.
Final Answer: (C)