If sin ^{-1}(4 x)-cos ^{-1}(3 x)=frac{pi}{6}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$....$(i)$Let $A=\sin ^{-1}(4 x)$ and $B=\cos ^{-1}(3 x)$.Then $\sin A=4 x$ and $\cos B=3 x$.

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2 \sqrt{13}}$

Vedclass · Answer

(C) Given: $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$....$(i)$Let $A=\sin ^{-1}(4 x)$ and $B=\cos ^{-1}(3 x)$.Then $\sin A=4 x$ and $\cos B=3 x$.We know that $\cos A=\sqrt{1-16 x^2}$ and $\sin B=\sqrt{1-9 x^2}$.From $(i)$, $A-B=\frac{\pi}{6}$. Taking sine on both sides:$\sin(A-B) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we get:$(4x)(3x) - \sqrt{1-16 x^2} \sqrt{1-9 x^2} = \frac{1}{2}$.$12x^2 - \frac{1}{2} = \sqrt{(1-16 x^2)(1-9 x^2)}$.Squaring both sides:$(12x^2 - \frac{1}{2})^2 = (1-16 x^2)(1-9 x^2)$.$144x^4 - 12x^2 + \frac{1}{4} = 1 - 9x^2 - 16x^2 + 144x^4$.$-12x^2 + \frac{1}{4} = 1 - 25x^2$.$13x^2 = 1 - \frac{1}{4} = \frac{3}{4}$.$x^2 = \frac{3}{52}$.$x = \sqrt{\frac{3}{52}} = \frac{\sqrt{3}}{2 \sqrt{13}}$.

If $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$, then $x=$

Similar Questions

$\tan^{-1} [2 \cos (2 \sin^{-1} \frac{1}{2})] = \dots \dots \dots$

Find the principal value of $cosec^{-1}(2)$.

The value of $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ is

The value of $\cos (\tan^{-1} x)$ is equal to : . . . . . . (Where $|x| < 1$)

$\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(-\frac{1}{2}\right)}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module