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Mathematics Question on Trigonometric Equations

If sin⁡ A+cos⁡ B=a\sin \, A + \cos \, B = a and sin⁡ B+cos⁡ A=b\sin \, B + \cos \, A = b, then sin⁡(A+B)\sin (A + B) is equal to

A

a2+b22\frac{a^2 + b^2}{2}

B

a2−b2+22\frac{a^2 - b^2 + 2 }{2}

C

a2+b2−22\frac{a^2 + b^2 - 2 }{2}

D

None of these

Answer

a2+b2−22\frac{a^2 + b^2 - 2 }{2}

Explanation

Solution

sin⁡A+cos⁡B=a…\sin A +\cos B = a \ldots (i)
sin⁡B+cos⁡A=b…\sin B +\cos A = b \ldots (ii)
Squaring and adding (i) and (ii)
sin⁡2A+cos⁡2B+2sin⁡Acos⁡B+sin⁡2B+cos⁡2A+2cos⁡Asin⁡B=a2+b2\sin ^{2} A +\cos ^{2} B +2 \sin A \cos B +\sin ^{2} B +\cos ^{2} A +2 \cos A \sin B = a ^{2}+ b ^{2}
sin⁡2A+cos⁡2A+sin⁡2B+cos⁡2B+2(sin⁡Acos⁡B+cos⁡Asin⁡B)=a2+b2\sin ^{2} A +\cos ^{2} A +\sin ^{2} B +\cos ^{2} B +2(\sin A \cos B +\cos A \sin B )= a ^{2}+ b ^{2}
1+1+2sin⁡(A+B)=a2+b21+1+2 \sin ( A + B )= a ^{2}+ b ^{2}
⇒sin⁡(A+B)=a2+b2−22\Rightarrow \sin ( A + B )=\frac{ a ^{2}+ b ^{2}-2}{2}