Question

Let \(t\in (0,1)\) and \(α\in (0,\frac{π}{4})\). If \(x = \text{cosec}^{-1}(\frac{1+t^2}{2t})\), \(y = cot^{-1}(\frac{\sqrt{1-t^2}}{t})\) and \(\frac{dy}{dx} = f(t)\), then the value of \(f(tanα)\) is

• A. \(\frac{secα}{\sqrt{cos2α}}\)
• B. \(\frac{secα}{2\sqrt{cos2α}}\)
• C. \(\frac{cosα}{2\sqrt{cos2α}}\)
• D. \(\frac{cosα}{\sqrt{cos2α}}\)

Hint:

Simplify x and y to simple inverse functions of t first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Convert \(x\) and \(y\) to simpler forms. Then \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Step 2: Key Formula or Approach:
\(\text{cosec}\,x = \frac{1 + t^2}{2t}\), so \(\sin x = \frac{2t}{1 + t^2}\), which means \(x = 2\tan^{-1}t\) for \(t \in (0,1)\). For \(y\), put \(t = \sin\theta\): \(\cot y = \frac{\cos\theta}{\sin\theta}\), so \(y = \theta = \sin^{-1}t\).

Step 3: Detailed Explanation:
\(\frac{dx}{dt} = \frac{2}{1 + t^2}\) and \(\frac{dy}{dt} = \frac{1}{\sqrt{1 - t^2}}\).
\[ f(t) = \frac{dy}{dx} = \frac{1 + t^2}{2\sqrt{1 - t^2}} \]
Now put \(t = \tan\alpha\): \(1 + t^2 = \sec^2\alpha\) and \(\sqrt{1 - \tan^2\alpha} = \frac{\sqrt{\cos 2\alpha}}{\cos\alpha}\).
\[ f(\tan\alpha) = \frac{\sec^2\alpha \cdot \cos\alpha}{2\sqrt{\cos 2\alpha}} = \frac{\sec\alpha}{2\sqrt{\cos 2\alpha}} \]

Final Answer:
\(f(\tan\alpha) = \frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}\), option (B). \[ \boxed{\frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}} \]