The value of tan14sin−1⁡638 is - Infinity Learn by Sri Chaitanya

A
$\frac{1}{\sqrt{7}}$
B
$\frac{2}{\sqrt{7}}$
C
$\sqrt{\frac{7}{8}}$
D
$\frac{1}{\sqrt{8}}$

Solution:

Let $\sin^{- 1} \frac{\sqrt{63}}{8} = θ .$ Then $, \sin θ = \frac{\sqrt{63}}{8}$ and $\cos θ = \frac{1}{8} .$

Now,

$\begin{array}{l} \cos θ = \frac{1}{8} \Rightarrow 2 \cos^{2} \frac{θ}{2} - 1 = \frac{1}{8} \Rightarrow \cos^{2} \frac{θ}{2} = \frac{9}{16} \Rightarrow \cos \frac{θ}{2} = \frac{3}{4} \\ ∴ \tan (\frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8}) = \tan \frac{θ}{4} = \sqrt{\frac{1 - \cos \frac{θ}{2}}{1 + \cos \frac{θ}{2}}} = \sqrt{\frac{1 - \frac{3}{4}}{1 + \frac{3}{4}}} = \frac{1}{\sqrt{7}} \end{array}$

The value of tan $(\frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8})$ is

Solution:

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