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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CUET UG Official Maths Paper (Held on_23 May 23)
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  1. \(\rm tan^{-1}\frac{x}{a}\)
  2. tan-1 (ax)
  3. \(\rm a \tan^{-1}\frac{x}{a}\)
  4. \(\rm \sin^{-1}\frac{x}{a}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \sin^{-1}\frac{x}{a}\)
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Detailed Solution

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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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