Question

If \[ y=\sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right), \quad x=\cos 2\theta, \] then \[ \frac{dy}{dx}= \]

• A. \(\dfrac{x}{\sqrt{1-x^2}}\)
• B. \(-\cot 2\theta\)
• C. \(\tan 2\theta\)
• D. \(\dfrac{-x}{2\sqrt{1-x^2}}\)

Hint:

When \(x=\cos 2\theta\), use \[ \sqrt{\frac{1-x}{1+x}}=\tan\theta. \] This simplifies inverse trigonometric expressions quickly.

Answer 1

The Correct Option is B

Step 1: Substitute \(x=\cos 2\theta\).
Given, \[ x=\cos 2\theta. \] Now, \[ \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}. \] Using identities, \[ 1-\cos 2\theta=2\sin^2\theta \] and \[ 1+\cos 2\theta=2\cos^2\theta. \] Therefore, \[ \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \tan\theta. \]

Step 2: Simplify \(y\).
So, \[ y=\sin\left(2\tan^{-1}(\tan\theta)\right). \] Hence, \[ y=\sin 2\theta. \]

Step 3: Differentiate \(y\) and \(x\) with respect to \(\theta\).
Since \[ y=\sin 2\theta, \] we get \[ \frac{dy}{d\theta}=2\cos 2\theta. \] Also, \[ x=\cos 2\theta, \] so \[ \frac{dx}{d\theta}=-2\sin 2\theta. \]

Step 4: Find \(\frac{dy}{dx}\).
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}. \] \[ = \frac{2\cos 2\theta}{-2\sin 2\theta}. \] \[ = -\cot 2\theta. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{-\cot 2\theta} \]