Question

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1​

Answer 1

Answer:

See below.

Step-by-step explanation:

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1​

The 2 halves of this expression are the identities for [sin(A + B)]^2 and

[(cos (A + B)] ^2 respectively , therefore:

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2 =  [sin(A + B)]^2 +

[(cos (A + B)] ^2

Using the identity sin^2Ф + cos^2Ф = 1  we see that if we put Ф = (A + B) we have

 [sin(A + B)]^2 + [(cos (A + B)] ^2 = 1  so the identity

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2 = 1  must be true also.