કિંમત શોધો: sin ^{-1}left(frac{3}{5}right) + | vedclass

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

Question

(A) ધારો કે $	heta = \sin ^{-1}\left(\frac{3}{5}ight)$. તેથી $\sin 	heta = \frac{3}{5}$.નિત્યસમ $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin ^

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(A) ધારો કે $	heta = \sin ^{-1}\left(\frac{3}{5}ight)$. તેથી $\sin 	heta = \frac{3}{5}$.નિત્યસમ $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin ^2 	heta}} = \frac{3/5}{\sqrt{1 - (3/5)^2}} = \frac{3/5}{4/5} = \frac{3}{4}$ નો ઉપયોગ કરતા.આમ,$\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.હવે,પદાવલિ $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{7}ight)$ બને છે.સૂત્ર $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ ($xy $	an ^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} 	imes \frac{1}{7}}ight) = 	an ^{-1}\left(\frac{\frac{21+4}{28}}{1 - \frac{3}{28}}ight) = 	an ^{-1}\left(\frac{25/28}{25/28}ight) = 	an ^{-1}(1) = \frac{\pi }{4}$.

કિંમત શોધો: $\sin ^{-1}\left(\frac{3}{5}\right) + \tan ^{-1}\left(\frac{1}{7}\right) = $

Similar Questions

સાબિત કરો કે $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$,જ્યાં $|x| < \frac{1}{\sqrt{3}}$.

$\cos \left(\tan ^{-1}\left(\sin \left(\cot ^{-1} x\right)\right)\right)$ નું મૂલ્ય શું છે?

જો ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{2}$ હોય,તો $x$ ની કિંમત શોધો.

$\sin ^{-1}\left(\cos \frac{\pi}{13}\right)+\cos ^{-1}\left(\sin \frac{\pi}{13}\right) = $ . . . . . . .

જો $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$ હોય,તો $x + y + z$ ની કિંમત શું થાય?

Vedclass Test Series

Exam Paper Generator

Online Exam Module