Write the function in the simplest form: \(\tan^{-1}\bigg(\frac{\cos x-\sin x}{\cos x+\sin x}\bigg)0<x<\pi\)
\(\tan^{-1}\bigg(\frac{\cos x-\sin x}{\cos x+\sin x}\bigg)\)
= \(\tan^{-1}\bigg(\frac{\frac{1-\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\bigg)\)
=\(\tan^{-1}\bigg(\frac{1-\tan x}{1+\tan x}\bigg)\)
=\(\tan^{-1}(1) -\tan^{-1}(\tan x)\:[\tan^{-1}\frac{x-y}{1-xy}=\tan^{-1}x-\tan^{-1}y]\)
= \(\frac{\pi}{4-x}\)
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)