If \(tan^{-1}\frac{x-1}{x-2}+tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4}\) then find the value of x
\(tan^{-1}\frac{x-1}{x-2}+tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4}\)
=>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
The equation \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] is valid for all values of \(x\) satisfying:
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)