Question

Given, cos(A+B)=cosA.cosB-sin A.sin B; prove that: sin2A+cos2A=1​

Answer 1

Answer:

HERE IS YOUR ANSWER DUDE

Explanation:

Given that: cos A + cos B = 0

Hence, cos B = - cos A —(1)

Given that: sin A + sin B = 0

Hence, sin B = - sin A — (2)

Now,

L.H.S. = cos 2A + cos 2B

= (cos^2 A - sin^2 A) + (cos^2 B - sin^2 B)

= (cos^2 A - sin^2 A) + ((- cos A)^2 - (-sin A)^2) — by substituting based on (1) & (2)

= (cos^2 A - sin^2 A) + (cos^2 A - sin^2 A)

= 2 (cos^2 A - sin^2 A)

= 2 (cos A . cos A - sin A . sin A)

= 2 (cos A . (-cos B) - sin A . (-sin B)) — by substituting based on (1) & (2)

= 2 (- cos A cos B + sin A sin B)

= -2(cos A cos B - sin A sin B)

= -2 cos (A+B)

= 2 cos (π+A+B)

= R.H.S.

HENCE LHS=RHS

HENCE PROVED

MARK AS BRAINLIEST HEHE

Answer 2

i hope its help you :-)......