Question

If \[ y = \cot^{-1}\!\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right), \quad \text{then find} \quad \frac{dy}{dx}. \]

• A. \(\dfrac{1}{2}\)
• B. \(-1\)
• C. \(\dfrac{1}{3}\)
• D. \(1\)

Hint:

Expressions like \(\sqrt{\frac{1-\sin x}{1+\sin x}}\) often simplify to \(\sec x-\tan x\), making differentiation easy.

Answer 1

The Correct Option is A

Step 1: Simplify the expression inside \(\cot^{-1}\).
Let \[ t=\sqrt{\frac{1-\sin x}{1+\sin x}} \] Now use the identity: \[ \frac{1-\sin x}{1+\sin x}=\left(\frac{1-\sin x}{\cos x}\right)^2 \] So, \[ t=\frac{1-\sin x}{\cos x}=\sec x-\tan x \]
Step 2: Convert \(\cot^{-1}\) to \(\tan^{-1}\) form.
\[ y=\cot^{-1}(t) \Rightarrow y=\tan^{-1}\left(\frac{1}{t}\right) \] \[ \frac{1}{t}=\frac{1}{\sec x-\tan x}=\sec x+\tan x \] So, \[ y=\tan^{-1}(\sec x+\tan x) \]
Step 3: Differentiate using the standard identity.
A known result is: \[ \frac{d}{dx}\Big(\tan^{-1}(\sec x+\tan x)\Big)=\frac{1}{2} \]
Step 4: Final conclusion.
Hence, \[ \frac{dy}{dx}=\frac{1}{2} \]