Question

prove that (1+cotA+tanA) (sinA-tanA)=secA/cosec^2A -cosecA/sec^2A​

Answer 1

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answer for the question

Answer 2

/* There is a mistake in the question. It must be like this */

$\red{ (1 + cot A + tan A )( sin A - cos A ) = \frac{sec A}{Cosec^{2} A } - \frac{Cosec A}{ Sec^{2} A } }$

$LHS =\red{ (1 + cot A + tan A )( sin A - cos A )}\\= \Big( 1 + \frac{cos A}{sin A } + \frac{sin A }{Cos A } \Big) \Big( \frac{1}{Cosec A } - \frac{ 1}{ secA} \Big) \\= \Big( 1 + \frac{sec A}{Cosec A } + \frac{sec A }{Cosec A } \Big) \Big( \frac{1}{Cosec A } - \frac{ 1}{ secA} \Big) \\= \Big( \frac{Sec ACosec A + Cosec^{2} + Sec^{2} A }{Sec A Cosec A } \Big) \Big( \frac{Sec A - Cosec A }{Sec A Cosec A } \Big) \\= \frac{ (secA - Cosec A )( Sec^{2} A + Sec A Cosec A + Cosec^{2} A ) }{Sec^{2} A Cosec ^{2} A } \\= \frac{ Sec^{3} A - Cosec^{3} A }{ Sec^{2} A Cosec^{2} A } \\= \frac{Sec^{3} A }{ Sec^{2} A Cosec^{2} A } - \frac{Cosec^{3} A }{ Sec^{2} A Cosec^{2} A } \\\green {= \frac{SecA }{ Cosec^{2} A } - \frac{Cosec A }{ Sec^{2} A }} \\= RHS$

$Hence \:proved .$

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