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 $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha$, then $\alpha=$

Prove $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x$,where $-\frac{1}{\sqrt{2}} \leq x \leq 1$.

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

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$,then $x^{100}+y^{100}+z^{100}=$

If $(\cos ^{-1} x)^2-(\sin ^{-1} x)^2 > 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module