Question

Sin40-cos70=root3cos80

Answer 1

LHS =sin40-cos70

=sin40-cos(90-20)

=sin40-sin20

now use formula
sinA-sinB=2cos(A+B)/2.sin(A-B)/2

hence.
sin40-sin20=2cos30.sin10

=2 x √3/2 x cos80
=√3cos80° =RHS

Answer 2

Answer:

Step-by-step explanation:

LHS= sin40 - cos70

= sin40 - cos(90-20)

= sin40 - sin20

(Using identity sinC - cosD = 2Cos(C+D/2). Sin(C-D/2), we get

2cos(60/2).sin(20/2)

=2cos30.sin10

=2cos30.sin(90-80)

=2×root3/2.cos80

=root3.cos80

Hence proved

I hope it helps