Browse Questions

# Prove that $tan \bigg( \frac{\pi}{4}+\frac{1}{2} cos^{-1}\frac{a}{b} \bigg)+tan \bigg(\frac{\pi}{4}-\frac{1}{2} cos^{-1}\frac{a}{b} \bigg) = \frac{2b}{a}$

Toolbox:
• Take $cos^{-1}\large\frac{a}{b}=\theta$$\Rightarrow\:$ $cos\theta=\large\frac{a}{b}$
• $tan(A \pm B) = \large\frac{tanA \pm tanB}{1 \mp tanAtanB}$
• $cos\theta = \large\frac{1-tan^2\large\frac{\theta}{2}}{1+tan^2\large\frac{\theta}{2}}$
Put $cos^{-1}\large\frac{a}{b}=\theta$$\Rightarrow\:$ $cos\theta=\large\frac{a}{b}$
Then,L.H.S = $tan \bigg[ \large\frac{\pi}{4}+\large\frac{\theta}{2} \bigg] + tan\bigg[ \large\frac{\pi}{4}-\large\frac{\theta}{2} \bigg]$
Using the formula of tan(A+B) and tan(A-B) by taking A=$\frac{\pi}{4}\:and\:B=\frac{\theta}{2}$ we have
$tan \bigg[ \large\frac{\pi}{4}+\large\frac{\theta}{2}\bigg]=\frac{tan\frac{\pi}{4}+tan\frac{\theta}{2}}{1-tan\frac{\pi}{4}.tan\frac{\theta}{2}}=\large \frac{1+tan\large\frac{\theta}{2}}{1-tan\large\frac{\theta}{2}}$
$tan \bigg[ \large\frac{\pi}{4}-\large\frac{\theta}{2}\bigg]=\frac{tan\frac{\pi}{4}-tan\frac{\theta}{2}}{1+tan\frac{\pi}{4}.tan\frac{\theta}{2}}=\large \frac{1-tan\large\frac{\theta}{2}}{1+tan\large\frac{\theta}{2}}$
Substituting the values in L.H.S. we get
$L.H.S. =\large \frac{1+tan\large\frac{\theta}{2}}{1-tan\large\frac{\theta}{2}}+\large\frac{1-tan\large\frac{\theta}{2}}{1+tan\large\frac{\theta}{2}}$
$=\large\frac{(1+tan\frac{\theta}{2})^2+\large(1-tan\frac{\theta}{2})^2}{1-\large\:tan^2\frac{\theta}{2}}$
$= \bigg( \large\frac{2+2tan^2\large\frac{\theta}{2}}{1-tan^2\large\frac{\theta}{2}} \bigg)$
$= 2\bigg( \large\frac{1+tan^2\large\frac{\theta}{2}}{1-tan^2\large\frac{\theta}{2}} \bigg)$
But we know that $cos\theta = \large\frac{1-tan^2\large\frac{\theta}{2}}{1+tan^2\large\frac{\theta}{2}}$
$\Rightarrow\:\frac{1}{ cos\theta} = \large\frac{1-tan^2\large\frac{\theta}{2}}{1+tan^2\large\frac{\theta}{2}}$
Therefore, L.H.S.becomes $= \large\frac{2}{cos\theta}=2\large\frac{b}{a}=R.H.S$
edited Mar 20, 2013