JEE PYQ: Inverse Trigonometric Functions Question 7
Question 7 - 2021 (24 Feb Shift 2)
A possible value of $\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$ is:
(1) $\frac{1}{2\sqrt{2}}$
(2) $\frac{1}{\sqrt{7}}$
(3) $\sqrt{7} - 1$
(4) $2\sqrt{2} - 1$
Show Answer
Answer: (2)
Solution
Let $\sin^{-1}\frac{\sqrt{63}}{8} = \theta$, so $\sin\theta = \frac{\sqrt{63}}{8}$, $\cos\theta = \frac{1}{8}$. Then $2\cos^2\frac{\theta}{2} - 1 = \frac{1}{8}$, so $\cos^2\frac{\theta}{2} = \frac{9}{16}$, $\cos\frac{\theta}{2} = \frac{3}{4}$. Using half-angle: $\frac{1-\tan^2\frac{\theta}{4}}{1+\tan^2\frac{\theta}{4}} = \frac{3}{4}$, so $\tan\frac{\theta}{4} = \frac{1}{\sqrt{7}}$.
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