Question

if A+B =90˙ prove that√ tan A tan B + tan A cot B/ sin A secB - sin²B/cos²A = tan A

Answer 1

HELLO DEAR,

given that:-

A+B=90°

A=( 90° - B)---------(1)
$\sqrt{ \frac{ \tan( \alpha ) \tan( \beta ) + \tan( \alpha ) \cot( \beta ) }{ \sin( \alpha ) \sec( \beta ) } - \frac{ { \sin }^{2} \alpha }{ \cos ^{2} \beta } } \\ = > \sqrt{ \frac{ \tan( 90 - \beta ) \tan( \beta ) + \tan( 9 0 - \beta ) \cot( \beta ) }{ \sin( 90 - \beta ) \sec( \beta ) } - \frac{ { \sin }^{2} (90 - \beta ) }{ \cos ^{2} \beta } }...............using(1) \\ = > \sqrt{ \frac{ \cot( \beta ) \tan( \beta ) + \cot( \beta ) \cot( \beta ) }{ \cos( \beta ) \sec( \beta ) } - \frac{ { cos }^{2} \beta }{ \cos ^{2} \beta } } \\ = > \sqrt{ \frac{ \frac{1}{ \tan( \beta ) } \times \tan( \beta ) + \cot ^{2} \beta }{ \frac{1}{ \sec( \beta ) } \times \sec( \beta ) } - 1 } \\ = > \sqrt{1 + \cot^{2} \beta - 1 } \\ > \sqrt{ { \cot }^{2} \beta } = \cot( \beta ) \\ = > \cot(90 - \alpha )......using(1) \\ = > \tan( \alpha )$
$I \: HOPE \: ITS \: HELP \: YOU \: DEAR, \\ THANKS$
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