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Question

Mathematics Question on Inverse Trigonometric Functions

Write the function in the simplest form: tan−1xa2−x2,∣x∣<atan^{-1}\frac{x}{\sqrt{a^2-x^2}},\mid x\mid<a

Answer

tan−1xa2−x2tan^{-1}\frac{x}{\sqrt{a^2-x^2}}

put x=a sin⁡θ\sin\theta
⇒xa=sin⁡θ⇒θ=sin⁡−1(xa)\Rightarrow\frac{x}{a}=\sin\theta\Rightarrow \theta=\sin^{-1}(\frac{x}{a})

∴tan⁡−1xa2−x2=tan⁡−1asin⁡θa2−a2sin⁡2θ\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(asin⁡θa1−sin⁡2θ)\tan^{-1}\bigg( \frac{a \sin\theta}{a\sqrt{1-\sin^2\theta}}\bigg)

=tan⁡−1(asin⁡θacos⁡θ)\tan^{-1}\bigg(\frac{a \sin\theta}{a \cos\theta}\bigg)

= tan⁡−1(tan⁡θ)=θ=sin⁡−1xa\tan^{-1}(\tan\theta)=\theta= \sin^{-1}\frac{x}{a}