Question

If $\tan \frac{\theta }{2}=\sqrt{\frac{a-b}{a+b}}\tan \frac{\phi }{2}$, then prove that $\cos \alpha =\frac{a\cos \phi +b}{a+b\cos \phi }$

Answer 1

Given, $\tan \frac{\theta }{2}=\sqrt{\frac{a-b}{a+b}}\tan \frac{\phi }{2}$
Now, $\cos \theta =\frac{1-{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\frac{1-\frac{a-b}{a+b}{\tan }^2\phi }{1+\frac{a-b}{a+b}{\tan }^2\phi }$
$=\frac{1-\frac{a-b}{a+b}\frac{{\sin }^2\frac{\phi }{2}}{{\cos }^2\frac{\phi }{2}}}{1+\frac{a-b}{a+b}\frac{{\sin }^2\frac{\phi }{2}}{{\cos }^2\frac{\phi }{2}}}$
$=\frac{(a+b){\cos }^2\phi -(a-b){\sin }^2\phi }{(a+b){\cos }^2\phi +(a-b){\sin }^2\phi }$
$=\frac{a({\cos }^2\frac{\phi }{2}-{\sin }^2\frac{\phi }{2})+b({\cos }^2\frac{\phi }{2}+{\sin }^2\frac{\phi }{2})}{a({\cos }^2\frac{\phi }{2}+{\sin }^2\frac{\phi }{2})+b({\cos }^2\frac{\phi }{2}-{\sin }^2\frac{\phi }{2})}$
$=\frac{a\cos \phi +b}{a+b\cos \phi }$
Ans: 4