Question

Find $m$ if $\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b})+\tan (\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b})=\frac{mb}{a}$

Answer 1

If $\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b})+\tan (\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b})-\frac{mb}{a}$ ----------(1)

L.H.S. $\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b})+\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b})$

$[Let{\cos }^{-1}\frac{a}{b}=\theta ,then\cos \theta =\frac{a}{b}and\frac{1}{2}{\cos }^{-1}\frac{a}{b}=\frac{\theta }{2}]$ ----------(A)

$\Rightarrow \tan (\frac{\pi }{4}+\frac{\theta }{2})+\tan (\frac{\pi }{4}-\frac{\theta }{2})$

$=\frac{\tan \frac{\pi }{4}+\tan \frac{\theta }{2}}{1-\tan \frac{\pi }{4}\tan \frac{\theta }{2}}+\frac{\tan \frac{\pi }{4}-\tan \frac{\theta }{2}}{1+\tan \frac{\pi }{4}\tan \frac{\theta }{2}}$

$=\frac{1+\tan \theta /2}{1-\tan \theta /2}+\frac{1-\tan \theta /2}{1+\tan \theta /2}$

$=\frac{(1+\tan \theta /2{)}^2+(1-\tan \theta /2{)}^2}{1-{\tan }^2\theta /2}$

$=\frac{1+{\tan }^2\theta /2+2\tan \theta /2+1+{\tan }^2\theta /2-2\tan \theta /2}{1-{\tan }^2\theta }/2$

$=\frac{2(1+{\tan }^2\theta /2)}{1-ta{n}^2\theta /2}=\frac{2({\sin }^2\theta /2+{\cos }^2\theta /2)}{-\sin \theta /2+{\cos }^2\theta /2}$

$=\frac{2}{\cos \theta }$

$\because \cos 2\theta ={\cos }^2{\theta }^2\theta -{\sin }^2\theta$

$\therefore \cos \theta ={\cos }^2\theta /2-{\sin }^2\theta /2$

$=\frac{2b}{a}$ [from equation (A)]

Compare with equation, we get $m=2$