Question

sin(90°-A) cos(90°-A)= tan A/1+tan^2A​

Answer 1

Answer:

Step-by-step explanation:

Hi ,

LHS = sin(90 - A ) * cos ( 90 - A )

= CosA sinA ----( 1 )

RHS = tanA / ( 1 + tan² A )

= tan A / sec² A

= ( SinA/cosA ) / ( 1/ cos² A )

= ( SinA cos² A ) / cosA

= SinAcosA ----( 2 )

From ( 1 ) and ( 2 ) ,

LHS = RHS

Answer 2

Step-by-step explanation:

I will use alpha instead of cap a as angle

$RHS \: = \frac{ \tan \alpha }{1 + { \tan }^{2} \alpha } = \frac{ \frac{ \sin \alpha }{ \cos \alpha } }{1 + { \frac{ { \sin}^{2} \alpha }{ { \cos}^{2} \alpha } } } \\ = \frac{ \frac{ \sin \: \alpha }{ \cos \alpha } }{ \frac{ { \cos}^{2} \alpha + { \sin }^{2} \alpha }{ { \cos }^{2} \alpha } } \\ = \frac{ \frac{ \sin \: \alpha }{ \cos \alpha } }{ \frac{ 1 }{ { \cos }^{2} \alpha } } = \frac{ \sin \alpha }{ \cos \alpha } \times { \cos}^{2} \alpha \\ = \sin \: \alpha \cos \alpha \\ = \cos( {90}^{0} - \alpha ) \sin( {90}^{0} - \alpha ) =LHS\\ since \: \\ sin( {90}^{0} - \alpha ) \: = \cos \alpha \: \\ and \\ cos( {90}^{0} - \alpha ) \: = \sin \alpha$