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: Substitute $t = \tan\alpha$ early:
Then $x = \text{cosec}^{-1}\frac{1 + \tan^2\alpha}{2\tan\alpha} = \text{cosec}^{-1}\frac{1}{\sin 2\alpha} = 2\alpha$ (since $2\alpha \in (0, \frac\pi2)$).

Step 2: Differentiate in alpha:
$y = \sin^{-1}(\tan\alpha)$ gives $\frac{dy}{d\alpha} = \frac{\sec^2\alpha}{\sqrt{1 - \tan^2\alpha}}$ and $\frac{dx}{d\alpha} = 2$.
So $\frac{dy}{dx} = \frac{\sec^2\alpha}{2\sqrt{1 - \tan^2\alpha}} = \frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}$.

Final Answer:
The value is $\frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}$, option (B). \[ \boxed{\frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}} \]