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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If ( ^ - 1 (1 - x)) = ( ^ - 1 (- x)) , then x is

If sin⁡(cot⁡−1(1−x))=cos⁡(tan⁡−1(−x))\sin \left(\cot^{- 1} \left(1 - x\right)\right) = \cos \left(\tan^{- 1} \left(- x\right)\right), then xx isIf ( ^ - 1 (1 - x)) = ( ^ - 1 (- x)) , then x isIf ( ^ - 1 (1 - x)) = ( ^ - 1 (- x)) , then x is

Options

  1. A11
  2. B12\dfrac{1}{2}
  3. C00
  4. D−12- \dfrac{1}{2}

Correct answer

B. 12\dfrac{1}{2}

Step-by-step solution

sin⁡(cot⁡−1(1−x))=cos⁡(tan⁡−1(−x))\sin \left(\cot^{- 1} \left(1 - x\right)\right) = \cos \left(\tan^{- 1} \left(- x\right)\right)
Let, cot⁡−1(1−x)=θ\cot^{- 1} \left(1 - x\right) = \theta
⇒cotθ=(1−x)⇒ cot \theta = \left(1 - x\right)
If ( ^ - 1 (1 - x)) = ( ^ - 1 (- x)) , then x is
⇒sinθ=12−2x+x2⇒ sin \theta = \dfrac{1}{\sqrt{2 - 2x + {x}^{2}}}
Let, tan⁡−1(−x)=ϕ\tan^{- 1} \left(- x\right) = ϕ
⇒tanϕ=−x⇒cosϕ=11+x2⇒ tan ϕ = - x ⇒ cos ϕ = \dfrac{1}{\sqrt{1 + {x}^{2}}}
If ( ^ - 1 (1 - x)) = ( ^ - 1 (- x)) , then x is
∴sinθ=cosϕ∴ sin \theta = cos ϕ
⇒12−2x+x2=11+x2⇒ \dfrac{1}{\sqrt{2 - 2x + {x}^{2}}} = \dfrac{1}{\sqrt{1 + {x}^{2}}}
⇒1+x2=2−2x+x2⇒ 1 + {x}^{2} = 2 - 2x + {x}^{2}
⇒x=12⇒ x = \dfrac{1}{2}
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