For 0<x<1, evaluate 1 + x 2 ⋅ [ x cos ( cot − 1 x ) + sin ( cot − 1 x ) ] 2 − 1 .

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MathematicsInverse Trigonometric FunctionsT inverse T propertyMedium2 minQB

MathematicsMediumsingle choice

For 0<x<1, evaluate $1 + x^{2} \cdot [x cos (cot^{- 1} x) + sin (cot^{- 1} x)]^{2} - 1$ .

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Answer:

Solution:

$cos (cot^{- 1} x) = x / 1 + x^{2}$ , $sin (cot^{- 1} x) = 1/ 1 + x^{2}$ ; their weighted sum is $1 + x^{2}$ . Thus the inner radical is $x^{2} = x$ $\since$ x>0, yielding x√(1+x²).

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:T inverse T property

⏱ 2mℹ️ Source: QB

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