Question

Prove that:tan^'1 (1/2)=pi/4 - 1/2 cos^-1 (4/5)

Answer 1

Answer:

$tan^{-1}\frac{1}{2}=\frac{\pi}{4}-\frac{1}{2}cos^{-1}\frac{4}{5}$

Step-by-step explanation:

$tan^{-1}\frac{1}{2}=\frac{\pi}{4}-\frac{1}{2}cos^{-1}\frac{4}{5}$

Proof:

$Take\\\\\frac{1}{2}cos^{-1}\frac{4}{5}=A\\\\cos^{-1}\frac{4}{5}=2A\\\\\frac{4}{5}=cos2A$

$cos2A=\frac{4}{5}\\\\\frac{1-tan^2A}{1+tan^2A}=\frac{4}{5}\\\\5-5tan^2A=4+4tan^2A\\\\9tan^2A=1\\\\tan^2A=\frac{1}{9}\\\\tanA=\frac{1}{3}$

$Now,\\\\tan(\frac{\pi}{4}-\frac{1}{2}cos^{-1}\frac{4}{5})\\\\=tan(\frac{\pi}{4}-A)\\\\=\frac{tan\frac{\pi}{4}-tanA}{1+tan\frac{\pi}{4}.tanA}\\\\=\frac{1-tanA}{1+tanA}\\\\=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}\\\\=\frac{\frac{3-1}{3}}{\frac{3+1}{3}}\\\\=\frac{\frac{2}{3}}{\frac{4}{3}}\\\\=\frac{2}{4}\\\\=\frac{1}{2}$

This implies

$tan(\frac{\pi}{4}-\frac{1}{2}cos^{-1}\frac{4}{5})=\frac{1}{2}\\\\\frac{\pi}{4}-\frac{1}{2}cos^{-1}\frac{4}{5}=tan^{-1}\frac{1}{2}$