Question

please solve this...
it's very very urgent​

Answer 1

LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y)

Let ( π/4 -x) = A

(π/4 -y) = B

Then,

LHS = cosA.cosB -sinA.sinB

But we Know,

cos(A + B) = cosA.cosB-sinA.sinB use this,

= cos(A + B)

= cos{(π/4 -x)+(π/4 -y)}

=cos(π/2 -(x +y)}

We know,

Cos(π/2 -∅) = sin∅ use this ,

= sin(x + y) = RHS

Answer 2

Answer:

Let (π/4-x)=A

(π/4-y)=B

Then,

LHS=cosA.cosB-sinA.sinB

But we know

cos(A+B)=cosA.cosB-sinA.sinB use this

=cos(A+B)

=cos{(π/4-x)+(π/4-y)}

=cos(π/2-(X+y)}

we know

cos(π/2-π)=sinπ use this,

=sin(X+y)=RHS