Question

The solutions of the equation \(\sin [2\cos ^{-1}\{\cot (2\tan ^{-1}x)\}]=0\) are

• A. \(\pm 3\)
• B. \(1\pm \sqrt{2}\)
• C. \(2\pm \sqrt{2}\)
• D. \(\pm \sqrt{2}\)

Hint:

Let the inner expression be an angle phi in [0, pi]; sin(2 phi) = 0 forces cot(2 tan^-1 x) to take special values, then use the cot 2A formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(\phi =\cos ^{-1}\{\cot (2\tan ^{-1}x)\}\) with \(0\le \phi \le \pi\).
\(\sin 2\phi =0\) gives \(\phi =0,\frac{\pi }{2},\pi\), so \(\cot (2\tan ^{-1}x)=1,0,-1\).

Step 2: Express cot in x
Put \(x=\tan \alpha\). Then \(\cot 2\alpha =\frac{1-x^{2}}{2x}\).

Step 3: Take the case cot = -1
\(\frac{1-x^{2}}{2x}=-1\Rightarrow 1-x^{2}=-2x\Rightarrow x^{2}-2x-1=0\).
So \(x=1\pm \sqrt{2}\).

Step 4: Compare the options
The roots \(1\pm \sqrt{2}\) appear among the choices, matching option (B).

Final Answer:
The solutions are \(x=1\pm \sqrt{2}\). \[ \boxed{\text{(B) }1\pm \sqrt{2}} \]