Question

Prove that sin²A cos²B-cos²A sin² B = sin²A-sin²B

Answer 1

L.H.S=sin²A cos²B-cos²A sin²B
=sin²A(1-sin²B)- [(1-sin²A)sin²B]
=sin²A-sin²A sin²B-[sin²B- sin²A sin²B]
=sin²A-sin²A sin²B-sin²B+ sin²A sin²B
=sin²A-sin²B
=R.H.S

Answer 2

$\sin^2A \cos^2B-\cos^2A \sin^2 B = \sin^2A-\sin^2B$, proved.

Step-by-step explanation:

To prove that, $\sin^2A \cos^2B-\cos^2A \sin^2 B = \sin^2A-\sin^2B$.

L.H.S. = $\sin^2A \cos^2B-\cos^2A \sin^2 B$

Using the trigonometric identity,

$\sin^2\theta+ \cos^2\theta$ = 1

⇒ $\cos^2\theta$ = 1 - $\sin^2\theta$

= $\sin^2A (1-\sin^2 B)-(1-\sin^2 A) \sin^2 B$

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

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

= L.H.S., proved.

Thus, $\sin^2A \cos^2B-\cos^2A \sin^2 B = \sin^2A-\sin^2B$, proved.