Question

prove the above question​

Answer 1

Answer:

There u go mark as brainilist :)

Answer 2

LHS

$= sin( {90}^{o} - x)cos( {90}^{o} - x) \\ = cosx \times sinx \\ \binom{since \: sin( {90}^{o} - x) = cosx }{and \: cos( {90}^{o} - x) = sinx}$

RHS

$= \frac{tanx}{1 + {cot}^{2}( {90}^{o} - x) } \\ = \frac{tanx}{1 + {tan}^{2} x} \\ \binom{since \: cot( {90}^{o} - x) = tan( {90}^{o} - x) }{} \\ \\ = \frac{tanx}{ {sec}^{2}x } \\ \binom{since \: 1 + {tan}^{2}x = {sec}^{2}x }{?} \\ \\ = \frac{ \frac{sinx}{cosx} }{ \frac{1}{ {cos}^{2}x } } \\ = ( \frac{sinx}{cosx} )( \frac{ {cos}^{2} x}{1}) \\ = sinx \times cosx$

= LHS

•.• LHS = RHS

.•.

$sin( {90}^{o} - x)cos( {90}^{o} - x) \\ = \frac{tanx}{1 + {cot}^{2}( {90}^{o} - x) }$

... Hence Proved!