Write the function in the simplest form: \(\tan^{-1}\bigg(\sqrt{\frac{1-\cos x}{1+\cos x}}\bigg),x<\pi\)
\(\tan^{-1}\bigg(\sqrt{\frac{1-\cos x}{1+\cos x}}\bigg),x<\pi\)
= \(tan^{-1}(\sqrt \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\)
=\(\tan^{-1}\bigg(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\bigg)=\tan^{-1}\bigg(tan\frac{x}{2}\bigg)\)
=\(\frac{x}{2}\)
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)