frac{tan ^{-1}(sqrt{3})-sec ^{-1}(-2)}{operat | vedclass

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

Question

(B) We know the principal values of the inverse trigonometric functions: $	an ^{-1}(\sqrt{3}) = \frac{\pi}{3}$ $\sec ^{-1}(-2) = \pi - \sec ^{-1}(2)

Vedclass · Accepted Answer

$-\frac{4}{5}$

Vedclass · Answer

(B) We know the principal values of the inverse trigonometric functions: $	an ^{-1}(\sqrt{3}) = \frac{\pi}{3}$ $\sec ^{-1}(-2) = \pi - \sec ^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ $\operatorname{cosec}^{-1}(-\sqrt{2}) = -\operatorname{cosec}^{-1}(\sqrt{2}) = -\frac{\pi}{4}$ $\cos ^{-1}\left(-\frac{1}{2}ight) = \pi - \cos ^{-1}\left(\frac{1}{2}ight) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ Substituting these values into the expression: $\frac{\frac{\pi}{3} - \frac{2\pi}{3}}{-\frac{\pi}{4} + \frac{2\pi}{3}} = \frac{-\frac{\pi}{3}}{\frac{-3\pi + 8\pi}{12}} = \frac{-\frac{\pi}{3}}{\frac{5\pi}{12}} = -\frac{\pi}{3} 	imes \frac{12}{5\pi} = -\frac{4}{5}$

$\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(-\frac{1}{2}\right)}=$

Similar Questions

If $\frac{\pi}{2} \le x \le \frac{3\pi}{2},$ then $\sin^{-1}(\sin x)$ is equal to

If the area of the region bounded by the parabola $y^2 = 4kx$ and the line $x = k$, (where $k > 0$) is $\frac{128}{3}$ sq. units, then the value of $\sin^{-1} \left( \frac{2}{k} \right)$ is equal to...

$\sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]=$ . . . . . . .

The value of $\sin^{-1}(\sin 10)$ is

If $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$,then $\sinh \alpha=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module