Prove that:(cos40-sin30)/sin60-cos50=tan50

Answer 1

remember that sinx = cos(90-x)
- property of complementary angles, "co-sine" vs sine

Your statement is only true if
(cos40-sin30)/(sin60-cos50) =tan50 , but not the way you typed it.

cos40 = sin50
and sin60 = cos30
so
LS = (sin50 - sin30)/(cos30 - cos50)

remember that sinx - siny = 2 sin((x-y)/2) cos((x+y)/2)
and cosx - cosy = 2sin((x+y)/2) sin((x-y)/2)

LS = [2sin10 cos40]/[2sin40 sin10]
= cos40/sin40
= cot40
= tan50 ---> by complementary angles, "co-tangent" vs tangent
= RS

Answer 2

Prove that (cos40-sin30)/sin60-cos50 =tan50

LS = (cos40-sin30)/sin60-cos50

=(cos40-sin(90-60))/sin60-cos(90-40)
=(cos40-cos60)/sin60-sin40
=2sin(40+60/2)sin(60-40/2)/2cos(60+40/2)sin(60-40/2)

=2sin50 sin10 /2cos50 sin10

=sin50/cos50

=tan50