Question

$\huge \underline{\bf Prove \: that}$

${ \tan}{ - 1}\frac{x}{y} - { \tan}^{ - 1} \frac{x + y}{x - y}= \frac{\pi}{4}$
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Answer 1

Correct Question :–

• Prove that –

$\\ \implies\bf{ \tan}^{ - 1}\dfrac{x}{y} - { \tan}^{ - 1} \dfrac{x-y}{x+y}= \dfrac{\pi}{4}$

ANSWER :–

TO PROVE :–

$\\ \implies\bf{ \tan}^{ - 1}\dfrac{x}{y} - { \tan}^{ - 1} \dfrac{x-y}{x+y}= \dfrac{\pi}{4}$

SOLUTION :–

$\\ \implies\bf{ \tan}^{ - 1}\dfrac{x}{y} - { \tan}^{ - 1} \dfrac{x-y}{x+y}= \dfrac{\pi}{4}$

• Let's take L.H.S. –

$\\ \: = \: \bf{ \tan}^{ - 1} \bigg(\dfrac{x}{y} \bigg) - { \tan}^{ - 1} \bigg(\dfrac{x-y}{x+y} \bigg)$

• We know that –

$\\ \implies \bf{ \tan}^{ - 1}(x)- { \tan}^{ - 1}(y) = { \tan}^{ - 1} \bigg( \dfrac{x - y}{1 + xy} \bigg)$

• So that –

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \left( \dfrac{ \dfrac{x}{y} - \dfrac{x + y}{x - y}}{1 + \dfrac{x}{y} .\dfrac{x-y}{x+y}} \right)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \left( \dfrac{ \dfrac{x(x + y) - y(x - y)}{y(x + y)}}{\dfrac{y(x + y) + x(x - y)}{y(x + y)}} \right)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \left( \dfrac{x(x - y) - y(x + y)}{y(x - y) + x(x + y)} \right)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \left( \dfrac{x(x + y) - y(x - y)}{y(x + y) + x(x - y)} \right)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \left( \dfrac{ {x}^{2} + xy- xy + {y}^{2} }{xy + {y}^{2} + {x}^{2} - xy } \right)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \bigg( \dfrac{ {x}^{2} +{y}^{2} }{{y}^{2} + {x}^{2} } \bigg)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1} \bigg( \dfrac{ {x}^{2} +{y}^{2} }{{x}^{2} + {y}^{2} } \bigg)$

$\\ \: \: \bf = \: \: { \tan}^{ - 1}(1)$

$\\ \: \: \bf = \: \: \dfrac{\pi}{4}$

$\\ \: \: \bf = \: \: R.H.S.$

$\\ \: \: \: \: \large { \underline{ \underline{\bf Hence \: \: Proved}}}$