tan^{-1} left[ frac{sqrt{2}}{sqrt{3}} cos lef | vedclass

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

Question

(C) We know that $\sin^{-1}(1/\sqrt{2}) = \pi/4$.Substituting this into the expression, we get $	an^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left(

Vedclass · Accepted Answer

$-\pi/6$

Vedclass · Answer

(C) We know that $\sin^{-1}(1/\sqrt{2}) = \pi/4$.Substituting this into the expression, we get $	an^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left( 5 \cdot \frac{\pi}{4} ight) ight]$.Since $\cos(5\pi/4) = \cos(\pi + \pi/4) = -\cos(\pi/4) = -1/\sqrt{2}$.Substituting this value, the expression becomes $	an^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cdot \left( -\frac{1}{\sqrt{2}} ight) ight]$.This simplifies to $	an^{-1} \left( -\frac{1}{\sqrt{3}} ight)$.Since $	an^{-1}(-x) = -	an^{-1}(x)$ and $	an^{-1}(1/\sqrt{3}) = \pi/6$, the final result is $-\pi/6$.Thus, option $C$ is correct.

$\tan^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left( 5 \sin^{-1} \frac{1}{\sqrt{2}} \right) \right] = \dots$

Similar Questions

$\tan ^{-1} \sqrt{3} - \sec ^{-1}(-2)$ is equal to

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

The principal value of $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ is

Value of $\tan ^{-1}\left(\frac{\sin 2-1}{\cos 2}\right)$ is

If $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$,then considering positive square roots,$x$ has the value $\qquad$

Vedclass Test Series

Exam Paper Generator

Online Exam Module