Simplify tan ^{-1}left[frac{a cos x-b sin x}{ | vedclass

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

Question

(C) Given expression: $	an ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}ight]$Divide the numerator and denominator by $b \cos x$:$= 	an ^

Vedclass · Accepted Answer

$	an ^{-1} \frac{a}{b}-x$

Vedclass · Answer

(C) Given expression: $	an ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}ight]$Divide the numerator and denominator by $b \cos x$:$= 	an ^{-1}\left[\frac{\frac{a \cos x}{b \cos x} - \frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x} + \frac{a \sin x}{b \cos x}}ight]$$= 	an ^{-1}\left[\frac{\frac{a}{b} - 	an x}{1 + \frac{a}{b} 	an x}ight]$Using the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$,where $A = \frac{a}{b}$ and $B = 	an x$:$= 	an ^{-1}\left(\frac{a}{b}ight) - 	an ^{-1}(	an x)$$= 	an ^{-1}\left(\frac{a}{b}ight) - x$

Simplify $\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right],$ if $\frac{a}{b} \tan x > -1$.

Similar Questions

$2{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = $

The derivative of $\tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right)$ with respect to $\sin^{-1}(x)$ is

The sum to infinite terms of the series $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{2}{9}\right) + \dots + \tan^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}\right) + \dots$ is

If $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ and $\beta = \tan^{-1}\left(\frac{1}{3}\right)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\alpha - \beta$ is equal to

If $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module