Question

If \( \sin\!\left(\tan^{-1}(x\sqrt2)\right)=\cot\!\left(\sin^{-1}\!\sqrt{1-x^2}\right),\; x\in(0,1) \), then the value of \(x\) is :

• A. \( \frac12 \)
• B. \( \frac13 \)
• C. \( \frac23 \)
• D. \( \frac58 \)

Answer 1

The Correct Option is C

Step 1: Understanding the Question
We are given an equation involving inverse trigonometric functions and we need to solve for \(x\) in the interval \( (0,1) \).
Step 2: Key Formula or Approach
The best approach is to convert the inverse trigonometric functions into algebraic expressions. We can do this by using right-angled triangles to represent the angles.
- If \( \theta = \tan^{-1}(p/b) \), then a triangle with opposite side \(p\) and adjacent side \(b\) can be used to find \( \sin\theta, \cos\theta \), etc. The hypotenuse would be \( \sqrt{p^2+b^2} \).
- Similarly, if \( \phi = \sin^{-1}(p/h) \), a triangle with opposite side \(p\) and hypotenuse \(h\) can be used. The adjacent side would be \( \sqrt{h^2-p^2} \).
Step 3: Detailed Explanation
Let's simplify the Left Hand Side (LHS) and Right Hand Side (RHS) separately.
LHS: \( \sin(\tan^{-1}(x\sqrt2)) \)
Let \( \theta = \tan^{-1}(x\sqrt2) \). This means \( \tan\theta = x\sqrt{2} = \frac{x\sqrt{2}}{1} \).
We can form a right-angled triangle with opposite side \( = x\sqrt{2} \) and adjacent side \( = 1 \).
The hypotenuse \( = \sqrt{(x\sqrt{2})^2 + 1^2} = \sqrt{2x^2 + 1} \).
Therefore, \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x\sqrt{2}}{\sqrt{1+2x^2}} \).
RHS: \( \cot(\sin^{-1}\sqrt{1-x^2}) \)
Let \( \phi = \sin^{-1}\sqrt{1-x^2} \). This means \( \sin\phi = \sqrt{1-x^2} = \frac{\sqrt{1-x^2}}{1} \).
We can form a right-angled triangle with opposite side \( = \sqrt{1-x^2} \) and hypotenuse \( = 1 \).
The adjacent side \( = \sqrt{1^2 - (\sqrt{1-x^2})^2} = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = x \) (since \(x>0\)).
Therefore, \( \cot\phi = \frac{\text{adjacent}}{\text{opposite}} = \frac{x}{\sqrt{1-x^2}} \).
Equating LHS and RHS:
\[ \frac{x\sqrt{2}}{\sqrt{1+2x^2}} = \frac{x}{\sqrt{1-x^2}} \] Since \( x \in (0,1) \), we know \( x \neq 0 \), so we can divide both sides by \(x\).
\[ \frac{\sqrt{2}}{\sqrt{1+2x^2}} = \frac{1}{\sqrt{1-x^2}} \] Square both sides to eliminate the square roots:
\[ \frac{2}{1+2x^2} = \frac{1}{1-x^2} \] Cross-multiply: \[ 2(1-x^2) = 1(1+2x^2) \] \[ 2 - 2x^2 = 1 + 2x^2 \] \[ 1 = 4x^2 \] \[ x^2 = \frac{1}{4} \] \[ x = \pm \frac{1}{2} \] Since we are given \( x \in (0,1) \), the solution is \( x = \frac{1}{2} \).
Step 4: Final Answer
The calculation yields \(x = 1/2\), which is option (A).