If sin ^{-1} x-cos ^{-1} 2 x=sin ^{-1}left(fr | vedclass

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

Question

(B) Given equation: $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$.Since $\sin ^{-1

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Given equation: $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}ight)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}ight)$.Since $\sin ^{-1}\left(\frac{\sqrt{3}}{2}ight)=\frac{\pi}{3}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}ight)=\frac{\pi}{6}$,we have $\sin ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$.Thus,$\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} 2 x$.Taking sine on both sides: $x=\sin\left(\frac{\pi}{6}+\cos ^{-1} 2 xight) = \sin\left(\frac{\pi}{6}ight)\cos\left(\cos ^{-1} 2 xight) + \cos\left(\frac{\pi}{6}ight)\sin\left(\cos ^{-1} 2 xight)$.$x = \frac{1}{2}(2x) + \frac{\sqrt{3}}{2}\sqrt{1-(2x)^2} = x + \frac{\sqrt{3}}{2}\sqrt{1-4x^2}$.This implies $\frac{\sqrt{3}}{2}\sqrt{1-4x^2} = 0$,so $1-4x^2=0$,which gives $x = \frac{1}{2}$ (as $x = -\frac{1}{2}$ does not satisfy the original equation).Now,calculate $	an ^{-1} x+	an ^{-1}\left(\frac{x}{x+1}ight)$ for $x=\frac{1}{2}$:$	an ^{-1}\left(\frac{1}{2}ight)+	an ^{-1}\left(\frac{1/2}{1/2+1}ight) = 	an ^{-1}\left(\frac{1}{2}ight)+	an ^{-1}\left(\frac{1}{3}ight)$.Using the formula $	an ^{-1} a + 	an ^{-1} b = 	an ^{-1}\left(\frac{a+b}{1-ab}ight)$:$	an ^{-1}\left(\frac{1/2+1/3}{1-(1/2)(1/3)}ight) = 	an ^{-1}\left(\frac{5/6}{5/6}ight) = 	an ^{-1}(1) = \frac{\pi}{4}$.

If $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$,then $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=$

Similar Questions

The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to $1$ are

If $y = \sum_{k=1}^{6} k \cos^{-1} \left\{ \frac{3}{5} \cos kx - \frac{4}{5} \sin kx \right\}$,then $\frac{dy}{dx}$ at $x = 0$ is

If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$,then $\frac{d y}{d x}$ is equal to

The number of solutions of $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = 0$ is

If $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$,then $\frac{dy}{dx}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module