MCQ
જો $\sin A + \sin B = C,\cos A + \cos B = D,$ તો $\sin (A + B) = $
- A$CD$
- B$\frac{{CD}}{{{C^2} + {D^2}}}$
- C$\frac{{{C^2} + {D^2}}}{{2\,CD}}$
- ✓$\frac{{2\,CD}}{{{C^2} + {D^2}}}$
$ \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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