Write the function in the simplest form: \(\tan^{-1}\frac{1}{\sqrt{x^2-1}},\mid{x}\mid>1\)
\(\tan^{-1}\frac{1}{\sqrt{x^2-1}},\mid{x}\mid>1\)
put x=cosec θ \(\Rightarrow\) θ= cosec-1 x
\(\tan^{-1} \frac{1}{\sqrt{x^2-1}}=\tan^{-1}\frac{1}{\sqrt{\cosec^2\theta-1}}\)
\(\tan^{-1}(\frac{1}{cot\theta})=\tan^{-1}(\tan\theta)\)
\(\Rightarrow \theta=\cosec^{-1}x=\frac{\pi} {2}-\sec^{-1}x\:[\cosec^{-1}x+\sec^{-1}x=\frac{\pi}{2}]\)
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)