Kerala KEAM 5 June 2024 Solved Paper

We are given the following: $$x = 5 \tan t \quad and \quad y = 5 \sec t.$$We need to find $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$.Step 1: Differentiate $x$ and $y$ with respect to $t$:$$\frac{dx}{dt} = \frac{d}{dt} \left( 5 \tan t \right) = 5 \sec^2 t.$$ $$\frac{dy}{dt} = \frac{d}{dt} \left( 5 \sec t \right) = 5 \sec t \tan t.$$Step 2: Use the chain rule to find $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{5 \sec t \tan t}{5 \sec^2 t} = \frac{\tan t}{\sec t}.$$Since $\frac{\tan t}{\sec t} = \sin t$, we have: $$\frac{dy}{dx} = \sin t.$$Step 3: Now, evaluate $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$: $$\sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}.$$Thus, the correct answer is option (C). Quick Tip: When given functions involving trigonometric identities, use differentiation rules such as the chain rule and standard trigonometric derivatives to simplify the process. In this case, use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ and trigonometric values for specific angles to evaluate the final result.