કિંમત શોધો: tan^{-1} left( frac{1 - x^2}{2x} | vedclass

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

Question

(B) ધારો કે $x = 	an 	heta$. તેથી $	heta = 	an^{-1} x$.આપેલ પદાવલિ $	an^{-1} \left( \frac{1 - x^2}{2x} ight) + \cos^{-1} \left( \frac{1 - x^2}{

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) ધારો કે $x = 	an 	heta$. તેથી $	heta = 	an^{-1} x$.આપેલ પદાવલિ $	an^{-1} \left( \frac{1 - x^2}{2x} ight) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$ છે.$x = 	an 	heta$ મૂકતા:$= 	an^{-1} \left( \frac{1 - 	an^2 	heta}{2 	an 	heta} ight) + \cos^{-1} \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight)$ત્રિકોણમિતીય નિત્યસમ $\cot 2	heta = \frac{1 - 	an^2 	heta}{2 	an 	heta}$ અને $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$ નો ઉપયોગ કરતા:$= 	an^{-1} (\cot 2	heta) + \cos^{-1} (\cos 2	heta)$કારણ કે $	an^{-1} (\cot 2	heta) = 	an^{-1} \left( 	an \left( \frac{\pi}{2} - 2	heta ight) ight) = \frac{\pi}{2} - 2	heta$ અને $\cos^{-1} (\cos 2	heta) = 2	heta$:$= \left( \frac{\pi}{2} - 2	heta ight) + 2	heta$$= \frac{\pi}{2}$.

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

Similar Questions

$\cot ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) = $ . . . . . .

જો $\theta = \sec^{-1}(\cosh u)$ હોય,તો $u =$

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ ની કિંમત શોધો.

$\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right) = $

જો $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ જ્યાં $a, x \in(0,1)$,તો $x$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module