Question

(iii) sin (90° - A) cos (90° - A) =
tan A/1+tan square A

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Answer 1

Step-by-step explanation:

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

tanA/1+tan^2A=tanA/sec A= sin A

• cosAsinA=sinA
• cosA=1
• A=90°

I hope this will help you

Answer 2

Step-by-step explanation:

GIVEN THE QUESTION:-

iii) sin (90° - A) cos (90° - A) =

tan A/1+tan square A

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