Question

A possible value of \( \tan\!\left(\frac{1}{4}\sin^{-1}\!\left(\frac{\sqrt{63}}{8}\right)\right) \) is :

• A. 1/√7
• B. 1/2√2
• C. √7 - 1
• D. 2√2 - 1

Hint:

Repeated use of the formula $\tan^2(\alpha/2) = \frac{1-\cos\alpha}{1+\cos\alpha}$ helps reduce the angle from $\theta$ to $\theta/2$ and then to $\theta/4$.

Answer 1

The Correct Option is A

Step 1: Let $\theta = \sin^{-1} \frac{\sqrt{63}}{8}$. Then $\sin \theta = \frac{\sqrt{63}}{8}$.
Step 2: $\cos \theta = \sqrt{1 - \frac{63}{64}} = \frac{1}{8}$.
Step 3: Use half-angle formula: $\tan^2 \frac{\theta}{2} = \frac{1-\cos\theta}{1+\cos\theta} = \frac{1-1/8}{1+1/8} = \frac{7}{9} \Rightarrow \tan \frac{\theta}{2} = \frac{\sqrt{7}}{3}$.
Step 4: Let $\phi = \frac{\theta}{2}$. We need $\tan \frac{\phi}{2}$.
Step 5: $\cos \phi$ from $\tan \phi = \frac{\sqrt{7}}{3}$ is $\frac{3}{4}$.
Step 6: $\tan^2 \frac{\phi}{2} = \frac{1-3/4}{1+3/4} = \frac{1/4}{7/4} = \frac{1}{7} \Rightarrow \tan \frac{\phi}{2} = \frac{1}{\sqrt{7}}$.