Question

If A + B + C = 2 S, then sin (S - A) sin (S - B) + sin S. sin (S - C) =

• A. Cos A cos B
• B. cos A cos B cos C
• C. Sin A sin B
• D. sin A sin B sin C

Answer 1

The correct option is C

Sin A sin B

sin (S - A) sin (S - B) + sin S. sin (S - C)

We can use formula 2 sin A sin B = cos (A - B) - cos (A + B)

So, multiply and divide given expression by 2

$\frac{1}{2}$ [2sin (S -A) sin(S-B) + 2 sin S.sin (S - C)]

= $\frac{1}{2}$ [cos (S - A - S + B) - cos (S - A + S - B) + cos (S - S + C) - cos (S + S - C)]

= $\frac{1}{2}$ [cos (-A + B) - cos (2S - A - B) + cosC - cos (2S - C)]

= $\frac{1}{2}$ [cos (-A + B) - cos (A + B + C - A - B) + cosC - cos (A + B + C - C)]

= $\frac{1}{2}$ [cos (-A + B) - cos C + cosC - cos (A + B)]

= $\frac{1}{2}$ [cos (-A + B) - cos (A + B)]

cos C - cos D = 2 sin $\frac{(C+D)}{2}$ . sin $\frac{(D-C)}{2}$

= $\frac{1}{2}$ $[2sin\left(\frac{-A+B+A+B}{2}\right).sin\left(\frac{A+B-(-A+B)}{2}\right)]$

= sin $\left(\frac{2B}{2}\right)$ . sin $\left(\frac{2A}{2}\right)$

= sin A . sin B