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Inverse Trigonometric Functions question

2017 · 9 Apr · Shift 1 · Q34
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Inverse Trigonometric Functions question

2017 · 9 Apr · Shift 1 · Q34

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
A value of x satisfying the equation sin[cot−1 (1+ x)] = cos [tan−1 x], is :
  1. A
    −12- {1 \over 2}−21​
  2. B
    −-− 1
  3. C
    0
  4. D
    12{1 \over 2}21​
View written solutionFree

Correct answer: A

  1. Given equation

We need to solve:

sin⁡(cot⁡−1(1+x))=cos⁡(tan⁡−1x)\sin\left(\cot^{-1}(1+x)\right)=\cos\left(\tan^{-1}x\right)sin(cot−1(1+x))=cos(tan−1x)

  1. Use standard inverse-trigonometric identities

Let

θ=cot⁡−1(1+x)\theta=\cot^{-1}(1+x)θ=cot−1(1+x)

Then

cot⁡θ=1+x\cot\theta=1+xcotθ=1+x

Using a right triangle, if cot⁡θ=adjacentopposite=1+x\cot\theta=\dfrac{\text{adjacent}}{\text{opposite}}=1+xcotθ=oppositeadjacent​=1+x, then

sin⁡θ=11+(1+x)2\sin\theta=\frac{1}{\sqrt{1+(1+x)^2}}sinθ=1+(1+x)2​1​

So,

sin⁡(cot⁡−1(1+x))=11+(1+x)2\sin\left(\cot^{-1}(1+x)\right)=\frac{1}{\sqrt{1+(1+x)^2}}sin(cot−1(1+x))=1+(1+x)2​1​

Now let

ϕ=tan⁡−1x\phi=\tan^{-1}xϕ=tan−1x

Then

tan⁡ϕ=x\tan\phi=xtanϕ=x

Using a right triangle,

cos⁡ϕ=11+x2\cos\phi=\frac{1}{\sqrt{1+x^2}}cosϕ=1+x2​1​

So,

cos⁡(tan⁡−1x)=11+x2\cos\left(\tan^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}cos(tan−1x)=1+x2​1​

  1. Equate both sides

Thus the equation becomes

11+(1+x)2=11+x2\frac{1}{\sqrt{1+(1+x)^2}}=\frac{1}{\sqrt{1+x^2}}1+(1+x)2​1​=1+x2​1​

Since both denominators are positive, we can equate them:

1+(1+x)2=1+x21+(1+x)^2=1+x^21+(1+x)2=1+x2

  1. Solve the algebraic equation

Expand:

1+(1+2x+x2)=1+x21+(1+2x+x^2)=1+x^21+(1+2x+x2)=1+x2

2+2x+x2=1+x22+2x+x^2=1+x^22+2x+x2=1+x2

Cancel x2x^2x2 from both sides:

2+2x=12+2x=12+2x=1

2x=−12x=-12x=−1

x=−12x=-\frac{1}{2}x=−21​

  1. Check with options

The value is:

x=−12x=-\frac{1}{2}x=−21​

So the correct option is A.

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