Write the function in the simplest form: \(tan^{-1}\frac{x}{\sqrt{a^2-x^2}},\mid x\mid<a\)
\(tan^{-1}\frac{x}{\sqrt{a^2-x^2}}\)
put x=a \(\sin\theta\)
\(\Rightarrow\frac{x}{a}=\sin\theta\Rightarrow \theta=\sin^{-1}(\frac{x}{a})\)
\(\therefore \tan^{-1}\frac{x}{\sqrt {a^2-x^2}}=\tan^{-1}\frac{a \sin\theta}{\sqrt {a^2-a^2\sin^2\theta}}\)
=\(\tan^{-1}\bigg( \frac{a \sin\theta}{a\sqrt{1-\sin^2\theta}}\bigg)\)
=\(\tan^{-1}\bigg(\frac{a \sin\theta}{a \cos\theta}\bigg)\)
= \(\tan^{-1}(\tan\theta)=\theta= \sin^{-1}\frac{x}{a}\)
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)