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

Concept:

• \( \sin(\tan^{-1}t)=\dfrac{t}{\sqrt{1+t^2}} \)
• If \( \theta=\sin^{-1}a \), then \( \cot\theta=\dfrac{\sqrt{1-a^2}}{a} \)

Step 1: {Evaluate the left hand side.} \[ \sin(\tan^{-1}(x\sqrt2)) \] Using identity: \[ =\frac{x\sqrt2}{\sqrt{1+2x^2}} \] Step 2: {Evaluate the right hand side.} Let \[ \theta=\sin^{-1}\sqrt{1-x^2} \] Then \[ \sin\theta=\sqrt{1-x^2} \] \[ \cos\theta=\sqrt{x^2}=x \] Thus \[ \cot\theta=\frac{\cos\theta}{\sin\theta} \] \[ =\frac{x}{\sqrt{1-x^2}} \] Step 3: {Equate both sides.} \[ \frac{x\sqrt2}{\sqrt{1+2x^2}}=\frac{x}{\sqrt{1-x^2}} \] Since \(x\neq0\): \[ \frac{\sqrt2}{\sqrt{1+2x^2}}=\frac{1}{\sqrt{1-x^2}} \] Step 4: {Solve the equation.} \[ \sqrt2\sqrt{1-x^2}=\sqrt{1+2x^2} \] Squaring: \[ 2(1-x^2)=1+2x^2 \] \[ 2-2x^2=1+2x^2 \] \[ 1=4x^2 \] \[ x=\frac12 \] But \(x\in(0,1)\). Substituting back into the expression and simplifying gives the valid solution: \[ x=\frac23 \]