Question

tan⁻¹(x/y)-tan⁻¹(x-y/x+y)=.......(x/y≥0),Select Proper option from the given options.
(a) π/4
(b) π/3
(c) π/2
(d) π

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

$\displaystyle\sf{ { \tan}^{ - 1} \bigg( \frac{x}{y} \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{x - y}{x + y} \bigg) \: }$

CALCULATION

$\displaystyle\sf{ { \tan}^{ - 1} \bigg( \frac{x}{y} \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{x - y}{x + y} \bigg) \: }$

$= \displaystyle\sf{ { \cot}^{ - 1} \bigg( \frac{y}{x} \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{1 - \frac{y}{x} }{1 + \frac{y}{x} } \bigg) \: }$

$\sf{Let \: \: \displaystyle\sf{ \frac{y}{x} } = { \tan} \theta \: \: \: \: \: so \: that \: \: \: \theta \: = { \tan}^{ - 1} \bigg( \frac{y}{x}\bigg) \: }$

Now

$\displaystyle\sf{ { \tan}^{ - 1} \bigg( \frac{x}{y} \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{x - y}{x + y} \bigg) \: }$

$= \displaystyle\sf{ { \cot}^{ - 1} \bigg( \frac{y}{x} \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{1 - \frac{y}{x} }{1 + \frac{y}{x} } \bigg) \: }$

$= \displaystyle\sf{ \frac{\pi}{2} - { \tan}^{ - 1} \bigg( \frac{y}{x} \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{1 - \frac{y}{x} }{1 + \frac{y}{x} } \bigg) \: }$

$= \displaystyle\sf{ \frac{\pi}{2} - { \tan}^{ - 1} \bigg( \tan \theta \bigg) - \:{ \tan}^{ - 1} \bigg( \frac{1 - \tan \theta }{1 + \tan \theta } \bigg) \: }$

$= \displaystyle\sf{ \frac{\pi}{2} - \theta- \:{ \tan}^{ - 1} \bigg( \frac{ \tan \frac{\pi}{4} - \tan \theta }{1 + \tan \frac{\pi}{4} \tan \theta } \bigg) \: }$

$= \displaystyle\sf{ \frac{\pi}{2} - \theta- \:{ \tan}^{ - 1} \tan \bigg( \frac{\pi}{4} - \theta\bigg) \: }$

$= \displaystyle\sf{ \frac{\pi}{2} - \theta- \: \bigg( \frac{\pi}{4} - \theta\bigg) \: }$

$= \displaystyle\sf{ \frac{\pi}{2} - \theta - \frac{\pi}{4} + \theta\: }$

$= \displaystyle\sf{ \frac{\pi}{2} - \frac{\pi}{4} \: }$

$= \displaystyle\sf{ \frac{\pi}{4} }$

RESULT

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