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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

The value of \ ^ - 1 [ ( ^ - 1 ((1)/(x)))]\ is equal to (x > 0)

The value of sin⁡{cot⁡−1[cos⁡(cot⁡−1(1x))]}\sin \left\{\cot^{- 1}\left[\cos \left(\cot^{- 1}\left(\dfrac{1}{x}\right)\right)\right]\right\} is equal to \left(x > 0\right)

Options

  1. A1+x22+x2\sqrt{\dfrac{1 + {x}^{2}}{2 + {x}^{2}}}
  2. B1−x22+x2\sqrt{\dfrac{1 - {x}^{2}}{2 + {x}^{2}}}
  3. C1+x22−x2\sqrt{\dfrac{1 + {x}^{2}}{2 - {x}^{2}}}
  4. D2+x21+x2\sqrt{\dfrac{2 + {x}^{2}}{1 + {x}^{2}}}

Correct answer

A. 1+x22+x2\sqrt{\dfrac{1 + {x}^{2}}{2 + {x}^{2}}}

Step-by-step solution

Let cot⁡−1(1x)=α⇒tanα=x\cot^{- 1}\left(\dfrac{1}{x}\right) = \alpha ⇒ tan \alpha\text{=}x
So, cosα=11+x2cos \alpha = \dfrac{1}{\sqrt{1 + {x}^{2}}}
∴sin⁡{cot⁡−1[cos⁡(tan⁡−1x)]}∴ \sin \left\{\cot^{- 1} \left[\cos \left(\tan^{- 1} x\right)\right]\right\}
=sin⁡(cot−111+x2)= \sin \left(co{t}^{- 1}\dfrac{1}{\sqrt{1 + {x}^{2}}}\right)
Let cot⁡−111+x2=β⇒cotβ=11+x2\cot^{- 1}\dfrac{1}{\sqrt{1 + {x}^{2}}} = \beta ⇒ cot \beta = \dfrac{1}{\sqrt{1 + {x}^{2}}}
∴sinβ=(1+x2)(2+x2)∴ sin \beta = \dfrac{\sqrt{( 1 + {x}^{2} )}}{\sqrt{( 2 + {x}^{2} )}}
∴sinβ=sin⁡{cot⁡−1[cos⁡(cot⁡−1(1x))]}=1+x22+x2∴ sin \beta = \sin \left\{\cot^{- 1}\left[\cos \left(\cot^{- 1} \left(\dfrac{1}{x}\right)\right)\right]\right\} = \sqrt{\dfrac{1 + {x}^{2}}{2 + {x}^{2}}}
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