sin^{-1}left(cos frac{53pi}{5}right) નું મૂલ્ | vedclass

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

Question

(D) આપેલ પદ: $\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$ આપણે $\frac{53\pi}{5}$ ને $10\pi + \frac{3\pi}{5}$ તરીકે લખી શકીએ છીએ. તેથી,$\cos\left(10\pi

Vedclass · Accepted Answer

$-\frac{\pi}{10}$

Vedclass · Answer

(D) આપેલ પદ: $\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$ આપણે $\frac{53\pi}{5}$ ને $10\pi + \frac{3\pi}{5}$ તરીકે લખી શકીએ છીએ. તેથી,$\cos\left(10\pi + \frac{3\pi}{5}\right) = \cos\left(\frac{3\pi}{5}\right)$. હવે,$\sin^{-1}\left(\cos \frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right)\right)$ $= \sin^{-1}\left(\sin\left(\frac{5\pi - 6\pi}{10}\right)\right)$ $= \sin^{-1}\left(\sin\left(-\frac{\pi}{10}\right)\right)$ $= -\frac{\pi}{10}$.

$\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$ નું મૂલ્ય શું છે?

Similar Questions

કિંમત શોધો: $\tan ^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)$

$\sec^{-1}[\sec(-30^o)] = $ ....... $^o$

જો $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$ હોય, તો $x=$

$\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right)$ ની કિંમત ....... $^o$ છે.

જો $x = \tan^{-1} \left(\frac{1}{5}\right) + \tan^{-1} \left(\frac{1}{8}\right)$ હોય, તો $\frac{\sin x + \cos x}{\tan x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module