Question

If $\sin \theta +\sin \varphi =a$ and $\cos \theta +\cos \varphi =b$, $(a\ne b,a\ne 0,b\ne 0)$ then -

• A. $\tan \theta +\tan \varphi =\frac{8ab}{(a^2+b^2{)}^2+4b^2}$
• B. $\cos \theta .\cos \varphi =\frac{(a^2+b^2{)}^2-4a^2}{4(a^2+b^2)}$
• C. $\cos (\theta +\varphi )=\frac{b^2-a^2}{b^2+a^2}$
• D. $\sin (\theta +\varphi )=\frac{4ab}{(a^2+b^2)+2b^2}$

Answer 1

Answer: (B), (C)

The correct options are
B $\cos \theta .\cos \varphi =\frac{(a^2+b^2{)}^2-4a^2}{4(a^2+b^2)}$
C $\cos (\theta +\varphi )=\frac{b^2-a^2}{b^2+a^2}$

Squaring and adding

$1+1+2\cos (\theta -\varphi )=a^2+b^2\Rightarrow \cos \left(\theta -\varphi \right)=\frac{a^2+b^2-2}{2}$


$2\sin \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)=a\underset{–}{2\cos \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)=b}\tan \left(\frac{\theta +\varphi }{2}\right)=\frac{a}{b}$

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

$\sin (\theta +\varphi )=\frac{2\tan \left(\frac{\theta +\varphi }{2}\right)}{1+{\tan }^2\left(\frac{\theta +\varphi }{2}\right)}=\frac{2\times \frac{a}{b}}{1+\frac{a^2}{b^2}}=\left(\frac{2ab}{a^2+b^2}\right)$

$\cos \theta \cos \varphi =\frac{1}{2}\left(\cos \right(\theta +\varphi )+\cos (\theta -\varphi \left)\right)=\frac{1}{2}(\frac{b^2-a^2}{b^2+a^2}+\frac{a^2+b^2-2}{2})$

$=\frac{(a^2+b^2{)}^2-4a^2}{4(a^2+b^2)}$

$\tan \theta +\tan \varphi =\frac{\sin \theta }{\cos \theta }+\frac{\sin \varphi }{\cos \varphi }=\frac{\sin (\theta +\varphi )}{\cos \theta .\cos \varphi }=\frac{8ab}{(a^2+b^2{)}^2-4a^2}$