Write the function in the simplest form: \(\tan ^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2}),a>0;\frac {-a}{\sqrt3} \leq x\leq\frac{ a}{\sqrt3}\)
\(\tan ^{-1}(\frac{3a^2x-x^3}{a^3-3ax^2})\)
put x= \(a \tan\theta\Rightarrow\frac{x}{a}=\tan\theta\Rightarrow\theta=\tan^{-1}\frac{x}{a}.\)
= \(\tan^{-1}\bigg(\frac{3a^2x-x^3}{a^3-3ax^2}\bigg)=\tan^{-1}\bigg(\frac{3a^2.a\tan\theta-a.3\tan\theta}{a^3-3a^3\tan^2\theta}\bigg)\)
=\(\tan^{-1}\bigg(\frac{3a^33\tan\theta-a.3\tan\theta}{a^3-3a^3\tan^2\theta}\bigg)\)
=\(\tan^{-1}\bigg(\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\bigg)=\tan^{-1}(\tan3\theta)\)
=3θ
= \(3\tan^{-1}\frac{x}{a}\)
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)