Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=2{\tan }^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2}\right)={\cos }^{-1}\left\{\frac{1-{\left(\sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2}\right)}^2}{1+{\left(\sqrt{\frac{a-b}{a+b}}\tan \frac{\theta }{2}\right)}^2}\right\}\left[\because 2{\tan }^{-1}\left(x\right)={\cos }^{-1}\left\{\frac{1-x^2}{1+x^2}\right\}\right] \\ ={\cos }^{-1}\left\{\frac{1-\frac{a-b}{a+b}{\tan }^2\frac{\theta }{2}}{1+\frac{a-b}{a+b}{\tan }^2\frac{\theta }{2}}\right\} \\ ={\cos }^{-1}\left\{\frac{a+b-\left(a-b\right){\tan }^2\frac{\theta }{2}}{a+b+\left(a-b\right){\tan }^2\frac{\theta }{2}}\right\} \\ ={\cos }^{-1}\left\{\frac{a+b-a{\tan }^2\frac{\theta }{2}+b{\tan }^2\frac{\theta }{2}}{a+b+a{\tan }^2\frac{\theta }{2}-b{\tan }^2\frac{\theta }{2}}\right\} \\ ={\cos }^{-1}\left\{\frac{a\left(1-{\tan }^2\frac{\theta }{2}\right)+b\left(1+{\tan }^2\frac{\theta }{2}\right)}{a\left(1+{\tan }^2\frac{\theta }{2}\right)+b\left(1-{\tan }^2\frac{\theta }{2}\right)}\right\} \\ ={\cos }^{-1}\left\{\frac{a{\displaystyle \left(\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)}+b\left({\displaystyle \frac{1+{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}\right)}{a\left({\displaystyle \frac{1+{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}\right)+b\left({\displaystyle \frac{1-{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}}\right)}\right\}\left[\mathrm{Dividing}{N}^r\mathrm{and}{D}^r\mathrm{by}1+{\tan }^2\frac{\theta }{2}\right] \\ ={\cos }^{-1}\left\{\frac{a{\displaystyle \left(\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)}+b}{a+b\left({\displaystyle \frac{1-{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}}\right)}\right\} \\ ={\cos }^{-1}\left\{\frac{a{\displaystyle \cos \theta }+b}{a+b\cos \theta }\right\}=\mathrm{RHS} \end{gathered}$$