The value of {tan ^{ - 1}}left( {frac{x}{y}} | vedclass

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

Question

(A) Given expression: ${	an ^{ - 1}}\left( {\frac{x}{y}} ight) - {	an ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} ight)$Divide the numerator and de

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(A) Given expression: ${	an ^{ - 1}}\left( {\frac{x}{y}} ight) - {	an ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} ight)$Divide the numerator and denominator of the second term by $x$:${	an ^{ - 1}}\left( {\frac{x}{y}} ight) - {	an ^{ - 1}}\left( {\frac{{1 - y/x}}{{1 + y/x}}} ight)$Using the formula ${	an ^{ - 1}}A - {	an ^{ - 1}}B = {	an ^{ - 1}}\left( {\frac{{A - B}}{{1 + AB}}} ight)$,we can write:${	an ^{ - 1}}\left( {\frac{{1 - y/x}}{{1 + 1 \cdot (y/x)}}} ight) = {	an ^{ - 1}}(1) - {	an ^{ - 1}}\left( {\frac{y}{x}} ight)$Substituting this back into the expression:${	an ^{ - 1}}\left( {\frac{x}{y}} ight) - \left( {\frac{\pi }{4} - {	an ^{ - 1}}\left( {\frac{y}{x}} ight)} ight)$Since ${	an ^{ - 1}}\left( {\frac{y}{x}} ight) = {\cot ^{ - 1}}\left( {\frac{x}{y}} ight)$:${	an ^{ - 1}}\left( {\frac{x}{y}} ight) + {\cot ^{ - 1}}\left( {\frac{x}{y}} ight) - \frac{\pi }{4}$Using the identity ${	an ^{ - 1}}	heta + {\cot ^{ - 1}}	heta = \frac{\pi }{2}$:$\frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}$.

The value of ${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right)$ is

Similar Questions

If $0 < x < 1$,then $\cot ^{-1}\left( \frac{2x^2 - 1}{2x\sqrt{1 - x^2}} \right)$ is equal to

$\cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right] = $

Prove that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

If $4\sin^{-1}x + \cos^{-1}x = \pi$,then $x$ is equal to

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2},$ then the value of $x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module