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Write the following in the simplest form: $ \tan^{-1} \large \frac{x}{\sqrt {a^2 - x^2}},$$ | x | < a $

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  • \( 1-sin^2\theta=cos^2\theta\)
Given $tan^{-1} \large \frac{x}{\sqrt {a^2 - x^2}}$,$ | x | < a$:
Let \( x=a\;sin\theta\) \( \Rightarrow \theta=sin^{-1}\large \frac{x}{a}\)
Also, we know that \( 1-sin^2\theta=cos^2\theta\)
Substituting for $x$ in the denominator, and reducing, we get \(\sqrt{a^2-x^2}=\sqrt{a^2-a^2\;sin^2\theta}\) \(=\sqrt{a^2(1-sin^2\theta)}=a\:cos\theta\)
The given expression reduces to \( tan^{-1}\bigg( \large \frac{a\: sin\theta}{a\: cos\theta} \bigg) \) \( = tan^{-1}\: tan\theta = \theta \)
Since \( x=a\;sin\theta\), the given expression reduces to \(\theta=sin^{-1}\large \frac{x}{a}\)
answered Feb 22, 2013 by thanvigandhi_1
edited Mar 14, 2013 by balaji.thirumalai

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