First slide

Properties of ITF

Question

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

Difficult

A

$\frac{1}{2 \sqrt{2}}$
B

$\sqrt{7} - 1$
C

$2 \sqrt{2} - 1$
D

$\frac{1}{\sqrt{7}}$

Solution

$\begin{matrix} \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} = θ \\ \sin 4 θ = \frac{\sqrt{63}}{8} \\ \frac{2 t a n 2 θ}{1 + \tan^{2} 2 θ} = \frac{\sqrt{63}}{8} \\ \sqrt{63} (1 + \tan^{2} 2 θ) = 16 \tan 2 θ \\ \sqrt{63} \tan^{2} 2 θ - 16 \tan 2 θ + \sqrt{63} = 0 \end{matrix}$
Use the quadratic formula for
It gives
$\begin{matrix} \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{9}{\sqrt{63}} o r \frac{7}{\sqrt{63}} \\ 7 (\tan^{2} θ - 1) + 2 \sqrt{63} \tan θ = 0 \\ \tan θ = \frac{- 2 \sqrt{63} \pm \sqrt{252 + 196}}{14} \\ = \frac{- 2 \sqrt{63} \pm \sqrt{448}}{14} \\ = \frac{- 6 \sqrt{7} \pm 8 \sqrt{7}}{14} \\ = \frac{1}{\sqrt{7}} \end{matrix}$

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