\(\rm \tan^{-1}\left\{\frac{x}{\sqrt{a^2-x^2}}\right\}\) का सरलतम रूप है, जहाँ -a < x < a है:

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This question was previously asked in

CUET UG Official Maths Paper (Held on_23 May 23)

Answer 1

स्पष्टीकरण:

माना x = a sin θ है, अर्थात θ = \(\rm \sin^{-1}\frac{x}{a}\)

तब,

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

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

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

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

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

= \(\rm \tan^{-1}(\tanθ)\)

= θ = \(\rm \sin^{-1}\frac{x}{a}\)

विकल्प (4) सत्य है।

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