cos left[ {{cos }^{ - 1}}left( {frac{{ - 1}}{ | vedclass

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

Question

(B) We know the identity for inverse trigonometric functions: $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ for $x \in [-1, 1]$.In the given expressio

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) We know the identity for inverse trigonometric functions: $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ for $x \in [-1, 1]$.In the given expression,let $x = -\frac{1}{7}$.Since $-\frac{1}{7} \in [-1, 1]$,we can apply the identity:$\cos^{-1}\left( -\frac{1}{7} \right) + \sin^{-1}\left( -\frac{1}{7} \right) = \frac{\pi}{2}$.Substituting this into the original expression:$\cos \left[ \cos^{-1}\left( -\frac{1}{7} \right) + \sin^{-1}\left( -\frac{1}{7} \right) \right] = \cos \left( \frac{\pi}{2} \right)$.Since $\cos \left( \frac{\pi}{2} \right) = 0$,the final result is $0$.

$\cos \left[ {{\cos }^{ - 1}}\left( {\frac{{ - 1}}{7}} \right) + {{\sin }^{ - 1}}\left( {\frac{{ - 1}}{7}} \right) \right] = $

Similar Questions

$\cos \left[ {{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2} \right] = $

Prove that $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$

If $\sin^{-1}(1 - x) - 2\sin^{-1}x = \pi/2$,then $x$ equals:

If $x$ takes non-positive permissible value,then $\sin^{-1} x =$

If in a triangle $ABC$,$A = \tan^{-1} 2$ and $B = \tan^{-1} 3$,then angle $C$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module