• # question_answer If $\sin \theta +\sin \varphi =a$and $\cos \theta +\cos \varphi =b,$then $\tan \frac{\theta -\varphi }{2}$is equal to [MP PET 1993] A) $\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{4-{{a}^{2}}-{{b}^{2}}}}$ B) $\sqrt{\frac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}$ C) $\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{4+{{a}^{2}}+{{b}^{2}}}}$ D) $\sqrt{\frac{4+{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}$

Given that $\sin \theta +\sin \varphi =a$ ?..(i) and $\cos \theta +\cos \varphi =b$ ?..(ii) Squaring, ${{\sin }^{2}}\theta +{{\sin }^{2}}\varphi +2\sin \theta \sin \varphi ={{a}^{2}}$ and ${{\cos }^{2}}\theta +{{\cos }^{2}}\varphi +2\cos \theta \cos \varphi ={{b}^{2}}$ Adding, 2+ 2 $(\sin \theta \sin \varphi +\cos \theta \cos \varphi )={{a}^{2}}+{{b}^{2}}$ Þ$2\cos (\theta -\varphi )={{a}^{2}}+{{b}^{2}}-2$Þ  $\cos (\theta -\varphi )=\frac{{{a}^{2}}+{{b}^{2}}-2}{2}$ $\Rightarrow \frac{1-{{\tan }^{2}}\frac{\theta -\varphi }{2}}{1+{{\tan }^{2}}\frac{\theta -\varphi }{2}}=\frac{{{a}^{2}}+{{b}^{2}}-2}{2}$ Þ $({{a}^{2}}+{{b}^{2}})+({{a}^{2}}+{{b}^{2}}){{\tan }^{2}}\frac{\theta -\varphi }{2}-2-2{{\tan }^{2}}\frac{\theta -\varphi }{2}$ $=2-2{{\tan }^{2}}\frac{\theta -\varphi }{2}$ Þ$\frac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}={{\tan }^{2}}\frac{\theta -\varphi }{2}$Þ $\tan \frac{(\theta -\varphi )}{2}=\sqrt{\frac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}$  Trick: Put$\theta =\frac{\pi }{2},\varphi ={{0}^{o}}$, then $a=1=b$ \$\tan \frac{\theta -\varphi }{2}=1$, which is given by (a) and (b). Again putting$\theta =\frac{\pi }{4}=\varphi$, we get$\tan \frac{\theta -\varphi }{2}=0$, which is given by (b).