tan left( 2{{cos }^{ - 1}}frac{3}{5} right) = | vedclass

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

Question

(C) ધારો કે $	heta = \cos^{-1}\left(\frac{3}{5}ight)$. તેથી $\cos 	heta = \frac{3}{5}$ થાય.$\cos 	heta = \frac{3}{5}$ હોવાથી,$\sin 	heta = \sqrt

Vedclass · Accepted Answer

$-\frac{24}{7}$

Vedclass · Answer

(C) ધારો કે $	heta = \cos^{-1}\left(\frac{3}{5}ight)$. તેથી $\cos 	heta = \frac{3}{5}$ થાય.$\cos 	heta = \frac{3}{5}$ હોવાથી,$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \left(\frac{3}{5}ight)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ મળે.આમ,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{4/5}{3/5} = \frac{4}{3}$ થાય.આપણે $	an(2	heta)$ ની કિંમત શોધવાની છે.સૂત્ર $	an(2	heta) = \frac{2	an 	heta}{1 - 	an^2 	heta}$ નો ઉપયોગ કરતા:$	an(2	heta) = \frac{2(4/3)}{1 - (4/3)^2} = \frac{8/3}{1 - 16/9} = \frac{8/3}{-7/9} = \frac{8}{3} 	imes \left(-\frac{9}{7}ight) = -\frac{24}{7}$.

$\tan \left( 2{{\cos }^{ - 1}}\frac{3}{5} \right) = $

Similar Questions

$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$ ની કિંમત શોધો.

$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2)$ ની કિંમત શોધો.

$1 + \cot^2(\sin^{-1}x) = $

જો $y = \sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$ અને $\frac{dy}{dx} = \frac{2^{x+1} \log 2}{f(x)}$ હોય,તો $f(0) = $ . . . . . .

$\tan \left( 90^o - \cot^{-1} \frac{1}{3} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module