A) \[CD\]
B) \[\frac{CD}{{{C}^{2}}+{{D}^{2}}}\]
C) \[\frac{{{C}^{2}}+{{D}^{2}}}{2\,CD}\]
D) \[\frac{2\,CD}{{{C}^{2}}+{{D}^{2}}}\]
Correct Answer: D
Solution :
As given \[\frac{\sin A+\sin B}{\cos A+\cos B}=\frac{C}{D}\] \[\Rightarrow \,\,\frac{2\,\,\sin \frac{A+B}{2}.\cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}.\cos \frac{A-B}{2}}=\frac{C}{D}\]\[\Rightarrow \,\,\tan \frac{A+B}{2}=\frac{C}{D}\] Thus, \[\sin \,(A+B)=\frac{2\,\,\tan \frac{A+B}{2}}{1+{{\tan }^{2}}\frac{A+B}{2}}\] \[=\frac{2\,\frac{C}{D}}{1+\frac{{{C}^{2}}}{{{D}^{2}}}}=\frac{2CD}{({{C}^{2}}+{{D}^{2}})}\].
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