tan left[frac{pi}{4}+frac{1}{2} cos ^{-1}left | vedclass

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

Question

(B) Let $	heta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)$,then $\cos 2 	heta = \frac{a}{b}$.The expression is $	an \left(\frac{\pi}{4} + 	h

Vedclass · Accepted Answer

$\frac{2 b}{a}$

Vedclass · Answer

(B) Let $	heta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)$,then $\cos 2 	heta = \frac{a}{b}$.The expression is $	an \left(\frac{\pi}{4} + 	hetaight) + 	an \left(\frac{\pi}{4} - 	hetaight)$.Using the identity $	an(A+B) + 	an(A-B) = \frac{2 	an A}{1 - 	an^2 A 	an^2 B}$ is not standard,so we expand:$= \left( \frac{1 + 	an 	heta}{1 - 	an 	heta} ight) + \left( \frac{1 - 	an 	heta}{1 + 	an 	heta} ight)$$= \frac{(1 + 	an 	heta)^2 + (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$$= \frac{1 + 2 	an 	heta + 	an^2 	heta + 1 - 2 	an 	heta + 	an^2 	heta}{1 - 	an^2 	heta}$$= \frac{2(1 + 	an^2 	heta)}{1 - 	an^2 	heta} = \frac{2}{\cos 2 	heta}$.Substituting $\cos 2 	heta = \frac{a}{b}$,we get $\frac{2}{a/b} = \frac{2b}{a}$.

$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

Similar Questions

If $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$,then $x=$ . . . . . . .

$\cot^{-1}(1) + \cot^{-1} (\frac{1}{2}) + \cot^{-1}(\frac{1}{3}) =$

$\sum\limits_{\lambda = 1}^{10} {{{\sin }^{ - 1}}\left( {\sin \left( {\lambda \pi - \frac{\pi }{6}} \right)} \right)} $ is equal to-

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = $

If $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$,then $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module