The simplest form of \(\rm \tan^{-1}\left\{\frac{x}{\sqrt{a^2-x^2}}\right\}\) is, where -a < x < a

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Question

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CUET UG Official Maths Paper (Held on_23 May 23)

Answer 1

Explanation:

Let x = a sin θ i.e., θ = \(\rm \sin^{-1}\frac{x}{a}\)

then

\(\rm \tan^{-1}\left\{\frac{x}{\sqrt{a^2-x^2}}\right\}\)

= \(\rm \tan^{-1}\left\{\frac{a\sinθ}{\sqrt{a^2-a^2\sin^2θ}}\right\}\)

= \(\rm \tan^{-1}\left\{\frac{a\sinθ}{\sqrt{a^2(1-\sin^2θ)}}\right\}\)

= \(\rm \tan^{-1}\left\{\frac{a\sinθ}{a\cosθ}\right\}\) (since sin²θ + cos²θ = 1)

= \(\rm \tan^{-1}\left\{\frac{\sinθ}{\cosθ}\right\}\)

= \(\rm \tan^{-1}(\tanθ)\)

= θ = \(\rm \sin^{-1}\frac{x}{a}\)

Option (4) is true.

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