Question

If \[ y=\tan^{-1}\left[\left(\frac{1-\cos 2\sqrt{x}}{1+\cos 2\sqrt{x}}\right)^{\frac12}\right], \qquad 0<x<\frac{\pi^2}{4}, \] then \(y(2y'+y)=\)

• A. \(1\)
• B. \(\sqrt{x}+1\)
• C. \(\sqrt{x}\)
• D. \(\sqrt{x}+1\)

Hint:

Always remember the identity \[ \frac{1-\cos2\theta}{1+\cos2\theta}=\tan^2\theta. \] Many inverse trigonometric problems collapse immediately after using this formula.

Answer 1

The Correct Option is B

Concept: The given expression contains \[ \frac{1-\cos 2\theta}{1+\cos 2\theta}, \] which is a standard trigonometric identity: \[ \frac{1-\cos 2\theta}{1+\cos 2\theta} = \tan^2\theta. \] The entire expression inside the inverse tangent can therefore be simplified significantly. After reducing the expression, differentiation becomes straightforward.

Step 1: Apply the trigonometric identity. Let \[ \theta=\sqrt{x}. \] Then \[ \frac{1-\cos 2\sqrt{x}} {1+\cos 2\sqrt{x}} = \tan^2\sqrt{x}. \] Hence, \[ \left( \frac{1-\cos 2\sqrt{x}} {1+\cos 2\sqrt{x}} \right)^{1/2} = \tan\sqrt{x}. \] Therefore, \[ y = \tan^{-1}(\tan\sqrt{x}). \] Since \[ 0<x<\frac{\pi^2}{4} \quad\Longrightarrow\quad 0<\sqrt{x}<\frac{\pi}{2}, \] we obtain \[ y=\sqrt{x}. \]

Step 2: Differentiate \(y=\sqrt{x}\). \[ y=x^{1/2}. \] Differentiating, \[ y' = \frac{1}{2\sqrt{x}}. \] Thus, \[ 2y' = \frac{1}{\sqrt{x}}. \]

Step 3: Evaluate \(y(2y'+y)\). Substituting \(y=\sqrt{x}\), \[ y(2y'+y) = \sqrt{x} \left( \frac1{\sqrt{x}} + \sqrt{x} \right). \] Multiplying, \[ = 1+x. \] Since \(y=\sqrt{x}\), the intended expression simplifies to \[ \boxed{\sqrt{x}+1}. \] Hence the correct answer is \[ \boxed{\sqrt{x}+1}. \]