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Question Number 17119 by gourav~ last updated on 01/Jul/17

Commented by prakash jain last updated on 01/Jul/17

$$\frac{\sin (A+3B)+\sin (3A+B)}{\sin 2A+\sin 2B}$$$$=\frac{2\sin \left(\frac{4A+4B}{2}\right)\cos \left(\frac{2B-2A}{2}\right)}{2\sin (A+B)\cos (A-B)}$$$$=\frac{2\sin \left(2(A+B)\right)\cos (A-B)}{2\sin (A+B)\cos (A-B)}$$$$=\frac{\sin \left(2(A+B)\right)}{\sin (A+B)}$$$$=\frac{2\sin (A+B)\cos (A+B)}{\sin (A+B)}$$$$=2\cos (A+B)◼$$

Commented by gourav~ last updated on 01/Jul/17

$$thankyousir$$

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