Question

If $\sin \theta +\sin \varphi =a,$ and $\cos \theta +\cos \varphi =b,$ then $\tan \frac{\theta -\varphi }{2}$ is equal to

• 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. $\sqrt{\frac{4+a^2+b^2}{a^2+b^2}}$

Answer 1

The correct option is B $\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$

$sin\theta +sin\varphi =a$ ________ (1) $cos\theta +cos\varphi =b$ _______ (2)

$(1{)}^2+(2{)}^2\Rightarrow a^2+b^2=si{n}^2\theta +si{n}^2\varphi +2sin\theta sin\varphi +co{s}^2\theta +co{s}^2\varphi +2cos\theta cos\varphi$

$(a^2+b^2)=1+1+2[sin\theta sin\varphi +cos\theta cos\varphi ]$

$(a^2+b^2)=2[1+cos(\theta -\varphi \left)\right]$

$\frac{a^2+b^2}{2}=2co{s}^2\left(\frac{\theta -\varphi }{2}\right)$

$\Rightarrow co{s}^2\left(\frac{\theta -\varphi }{2}\right)=\frac{a^2+b^2}{4}$ ______ (2)

$tan\left(\frac{\theta -\varphi }{2}\right)=\frac{\sqrt{1-co{s}^2\left(\frac{\theta -\varphi }{2}\right)}}{\sqrt{co{s}^2\left(\frac{\theta -\varphi }{2}\right)}}=\sqrt{\frac{1-\left(\frac{a^2+b^2}{4}\right)}{\left(\frac{a^2+b^2}{4}\right)}}=\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$

$\therefore tan\left(\frac{\theta -\varphi }{2}\right)=\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$