Question

Merhaba plzzzz solve neatly​

Answer 1

$\bold{To \: prove} = > \\ {tan}^{ - 1} ( \dfrac{1 - x}{1 + x} ) - {tan}^{ - 1} ( \dfrac{1 - y}{1 + y} ) = {sin}^{ - 1} ( \: \dfrac{y - x}{ \sqrt{1 + {x}^{2} } \sqrt{1 + {y}^{2} } } \: )$

${Concept \: used \\1) {tan}^{ - 1} x - {tan}^{ - 1} y = {tan}^{ - 1} ( \: \dfrac{x - y}{1 + xy} \: )}$

$2) \: if \: \\ {tan}^{ - 1} ( \dfrac{y + x}{1 - xy} ) = \alpha$

$= > tan \alpha = \dfrac{y + x}{1 - xy}$

$= > cot \alpha = \dfrac{y + x}{1 - xy}$

$= > {cot}^{2} \alpha = \dfrac{(y + x)^{2} }{ {(1 - xy)}^{2} }$

$= > {cosec}^{2} \alpha - 1 = \dfrac{ {(x + y)}^{2} }{ {(1 - xy)}^{2} }$

$= > {cosec}^{2} \alpha = 1 \: + \dfrac{ {(x + y)}^{2} }{ {(1 - xy)}^{2} }$

$= > {cosec}^{2} \alpha = \dfrac{ {(x + y)}^{2} + {(1 - xy)}^{2} }{ {(1 - xy)}^{2} }$

$= > {cosec}^{2} \alpha = \dfrac{ {x}^{2} + {y}^{2} + 2xy + 1 + {x}^{2} {y}^{2} - 2xy }{(1 - xy) ^{2} }$

$= > {cosec}^{2} \alpha = \dfrac{ {x}^{2} + {y}^{2} + 2xy + 1 + {x}^{2} {y}^{2} - 2xy }{(1 - xy) ^{2} }$

$= > {cosec}^{2} \alpha = \dfrac{1 + {x}^{2} + {y}^{2} + {x}^{2} {y}^{2} }{ {(1 - xy)}^{2} }$

$= > {cosec}^{2} \alpha = \dfrac{1 + {x}^{2} + {y}^{2}(1 + {x}^{2} ) }{ {(1 - xy)}^{2} }$

$= > {cosec}^{2} \alpha = \dfrac{(1 + {x}^{2}) \: (1 + {y}^{2}) }{ {(1 - xy)}^{2} }$

$= > {sin}^{2} \alpha = \dfrac{ {(1 - xy)}^{2} }{(1 + {x}^{2} ) \: (1 + {y}^{2}) }$

$= > sin \alpha = \dfrac{(1 - xy)}{ \sqrt{1 + {x}^{2} } \sqrt{1 + {y}^{2} } }$

$= > \alpha = {sin}^{ - 1} \dfrac{(1 - xy)}{ \sqrt{1 + {x}^{2} } \sqrt{1 + {y}^{2} } }$

$= >\boxed{\bold { {tan}^{ - 1} ( \dfrac{y + x}{1 - xy} ) = {sin}^{ - 1} ( \dfrac{1 - xy}{ \sqrt{1 + {x}^{2} } \sqrt{1 + {y}^{2} } } )}}$

$we \: use \: this \: relation \: later$

$\bold{Solution} = > LHS \\ {tan}^{ - 1} ( \dfrac{1 - x}{1 + x} ) - {tan}^{ - 1} ( \dfrac{1 - y}{1 + y} )$

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

$let \: x \: = tan \beta \: \: = > \beta = tan ^{ - 1}x \\ y = tan \gamma = > \gamma = {tan}^{ - 1} y \\ = {tan}^{ - 1} ( \: \dfrac{tan \dfrac{\pi}{4} - tan \beta }{1 + tan \dfrac{\pi}{4} tan \beta } \: ) \: - {tan}^{ - 1} ( \: \dfrac{tan \dfrac{\pi}{4} - tan \gamma }{1 + tan \dfrac{\pi}{4}tan \gamma } \: )$

$= > {tan}^{ - 1} \: tan \: ( \dfrac{\pi}{4} - \beta ) - {tan}^{ - 1} \: tan( \dfrac{\pi}{4} - \gamma )$

$= \dfrac{\pi}{4} - \beta - ( \dfrac{\pi}{4} - \gamma )$

$= \dfrac{\pi}{4} - \beta - \dfrac{\pi}{4} + \gamma$

$= \gamma \: - \: \beta$

$= {tan}^{ - 1} y \: - \: {tan}^{ - 1}x$

$= {tan}^{ - 1} \dfrac{y + x}{1 - xy}$

$applying \: abov e \: relation \: we \: get$

$= {sin}^{ - 1} ( \: \dfrac{1 - xy}{ \sqrt{1 + {x}^{2} } \sqrt{1 + {y}^{2} } } \: )$

$= RHS$

Answer 2

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