Write the function in the simplest form: \(\tan^{-1}\frac{\sqrt1+x^2-1}{x},x\neq0.\)
\(\tan^{-1}\frac{\sqrt1+x^2-1}{x}\),
put x=\(\tan\theta\)\(\Rightarrow\) \(\theta\)= \(\tan^{-1}x\)
\(\therefore \tan^{-1}\frac{\sqrt1+x^2-1}{x}\) =\(\tan^{-1}\frac{\sqrt1+\tan2\theta-1}{\tan\theta}\)
= \(\tan^{-1}\frac{(\sec\theta-1)}{\tan\theta}=\tan^{-1}(\frac{1-\cos\theta}{\sin\theta})\)
=\(\tan^{-1}\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}2\cos\frac{\theta}{2}}\)
=\(\tan^{-1}(\frac{\tan\theta}{2})=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x\)
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)