Question

Evaluate:

$\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}]=$ ?

Answer 1

$Letusassume,I=\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}]and\frac{1}{2}{\cos }^{-1}\frac{a}{b}=\theta \Rightarrow \cos 2\theta =\frac{a}{b}Thus,bysubstitutingthesetwovaluesinIweget,I=\tan [\frac{\pi }{4}+\theta ]+\tan [\frac{\pi }{4}-\theta ]=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=\frac{{(1+\tan \theta )}^2+{(1-\tan \theta )}^2}{1-{\tan }^2\theta }=\frac{2+2{\tan }^2\theta }{1-{\tan }^2\theta }=\frac{2(1+{\tan }^2\theta )}{2-{\sec }^2\theta }=\frac{2{\sec }^2\theta }{2-{\sec }^2\theta }=2\sec 2\theta =\frac{2}{\cos 2\theta }=\frac{2}{\frac{a}{b}}=\frac{2b}{a}.$