First slide

Properties of ITF

Question

If $\sin^{- 1} (6 x) + \sin^{- 1} (6 \sqrt{3} x) = - \frac{π}{2}$ then the value of $x$ is

Difficult

A

$\frac{1}{12}$
B

$- \frac{1}{12}$
C

$- \frac{1}{4 \sqrt{3}}$
D

$\frac{1}{4 \sqrt{3}}$

Solution

We have $\sin^{- 1} (6 x) + \sin^{- 1} (6 \sqrt{3} x) = - \frac{π}{2}$
$\Rightarrow \sin^{- 1} 6 x + \frac{π}{2} - \cos^{- 1} (6 \sqrt{3} x) = - \frac{π}{2} \Rightarrow π - \cos^{- 1} (6 \sqrt{3} x) = - \sin^{- 1} 6 x \Rightarrow \cos^{- 1} (- 6 \sqrt{3} x) = \sin^{- 1} (- 6 x) Let \cos^{- 1} (- 6 \sqrt{3} x) = \sin^{- 1} (- 6 x) = θ . \Rightarrow cosθ = - 6 \sqrt{3} x and sinθ = - 6 x ∴ \frac{sinθ}{cosθ} = \frac{- 6 x}{- 6 \sqrt{3} x} \Rightarrow tanθ = \frac{1}{\sqrt{3}} \Rightarrow θ = \frac{π}{6} ∴ sinθ = \sin \frac{π}{6} = \frac{1}{2} = - 6 x \Rightarrow x = - \frac{1}{12}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App