Question

prove that sin square A + sin square A minus b minus 2 sin a cos b Sin A minus b equal to sin square B

Answer 1

Answer:

proved

Step-by-step explanation:

Given  

prove that sin square A + sin square A minus b minus 2 sin a cos b Sin A minus b equal to sin square B

Given sin^2A + sin^2(A – B) – 2 sin A cos B sin(A – B) = sin^2 B

Consider the left hand side and taking common we get

Sin^2 A + sin (A – B) [sin(A – B) – 2 sinA cos B]

                               [sin A cos B – cos A sin B – 2 sin A cos B]

Sin^2 A – sin (A – B) [ sin A cos B + cos A sin B]

Sin^2 A – sin (A – B) sin (A + B)

We know that sin (A – B) sin (A + B )= (sin A cos B – cos A sin B)(sin A cos B + cos A sin B

Sin^2 A cos ^2 B – cos^2 A sin^2 B

Sin^2 A cos^2 B – (1 – sin^2 A)sin^2 B

Sin^2 A cos^2 B – sin^2 B + sin^2 Asin^2 B

Sin^2 A (cos ^2 B + sin^2 B) – sin^2 B

sin (A – B) sin (A + B ) = Sin^2 A – sin^2 B

Sin^2 A – sin^2 A + sin^2 B

Sin^2 B (proved)

Answer 2

Consider the provided information.

\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B

Consider the LHS.

\sin^2A\cos^2B-\cos^2A\sin^2B

\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B               (∴\cos^2x=1-\sin^2x)

\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B

\sin^2A-\sin^2B

Hence, proved.