tan ^{-1}left(frac{1+sqrt{3}}{3+sqrt{3}}right | vedclass

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

Question

(C) ધારો કે $x = 	an ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}ight)$.દલીલનું સાદું રૂપ આપતા: $\frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sq

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) ધારો કે $x = 	an ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}ight)$.દલીલનું સાદું રૂપ આપતા: $\frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$.તેથી,$x = 	an ^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$.ધારો કે $y = \sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}ight)$.દલીલનું સાદું રૂપ આપતા: $\sqrt{\frac{4(2+\sqrt{3})}{3(2+\sqrt{3})}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.તેથી,$y = \sec ^{-1}\left(\frac{2}{\sqrt{3}}ight) = \frac{\pi}{6}$.આમ,$x + y = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

$\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)$ ની કિંમત $.........$ છે.

Similar Questions

જો $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$ હોય,તો $\sinh \alpha=$

$\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ નું મુખ્ય મૂલ્ય શોધો.

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

$\sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=$

$\tan ^{-1}\left(\tan \frac{31 \pi}{6}\right)=$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module