Question

If $tan^{-1}\frac{x-1}{x-2}+tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4}$ then find the value of x

Answer 1

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

=>tan-1[x-1/x-2+x+1/x+2/1-(x-1/x-2+x+1/x+2)]=π/4

=>tan-1[(x-1)(x+2)+(x+1)(x-2)/(x+2)(x-2)-(x-1)(x+1)]=π/4

=>tan-1[x2+x-2+x2-x-2/x2-4-x2+1]=π/4
=>tan-1[2x2-4/-3]=π/4

tan(tan-1 4-2x2/3]=tanπ/4

=>4-2x2/3=1

=>4-2x2=3

2x2=4-3=1

x=±1/√2

Hence, the value of x is±1/√2