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 2θ to simplify the square root.
√[(1 - x)/(1 + x)] = √[(1 - cos 2θ)/(1 + cos 2θ)] = √[(2 sin²θ)/(2 cos²θ)] = tan θ.

Step 2: Simplify y.
y = sin(2 tan⁻¹(tan θ)) = sin(2θ).

Step 3: Differentiate y and x with respect to θ.
dy/dθ = 2 cos 2θ. dx/dθ = -2 sin 2θ.

Step 4: Compute dy/dx.
dy/dx = (dy/dθ)/(dx/dθ) = (2 cos 2θ)/(-2 sin 2θ) = -cot 2θ.

Step 5: Final conclusion.
The derivative is -cot 2θ.