Question

cos (90 - A) · sin (90 - A)/ tan (90 - A)= sin²A,Prove it.

Answer 1

we know, cos(90° - θ) = sinθ , sin(90° - θ) = cosθ
and tan(90° - θ) = cotθ .

now from above concepts
cos(90° - A) = sinA ------(1)
sin(90° - A) = cosA ------(2)
tan(90° - A) = cotA ------(3)

LHS = cos(90 - A).sin(90 - A)/tan(90 - A)

= sinA.cosA/cotA [ from (1), (2) and (3)]

= sinA.cosA/(cosA/sinA)

= sin²A = RHS

Answer 2

HELLO DEAR,


we know
sin(90° - A) = cosA
cos(90° - A) = sinA
tan(90° - A) = cotA

LHS = cos(90 - A).sin(90 - A)/tan(90 - A)

= sinA.cosA/cotA

= sinA.cosA/(cosA/sinA)

= sin²A = RHS

HENCE, L.H.S. = R.H.S.

I HOPE ITS HELP YOU DEAR,
THANKS