If sec^{-1} left( frac{x^2 + y^2}{x^2 - y^2} | vedclass

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(C) Given $\sec^{-1} \left( \frac{x^2 + y^2}{x^2 - y^2} \right) = 2a$, we have $\frac{x^2 + y^2}{x^2 - y^2} = \sec(2a)$.Applying componendo and divide

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$3$

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(C) Given $\sec^{-1} \left( \frac{x^2 + y^2}{x^2 - y^2} ight) = 2a$, we have $\frac{x^2 + y^2}{x^2 - y^2} = \sec(2a)$.Applying componendo and dividendo: $\frac{(x^2 + y^2) + (x^2 - y^2)}{(x^2 + y^2) - (x^2 - y^2)} = \frac{\sec(2a) + 1}{\sec(2a) - 1}$.$\frac{2x^2}{2y^2} = \frac{\frac{1}{\cos(2a)} + 1}{\frac{1}{\cos(2a)} - 1} = \frac{1 + \cos(2a)}{1 - \cos(2a)} = \frac{2\cos^2(a)}{2\sin^2(a)} = \cot^2(a)$.So, $\frac{x^2}{y^2} = \cot^2(a) \implies x^2 = y^2 \cot^2(a)$.Differentiating with respect to $x$: $2x = 2y \frac{dy}{dx} \cot^2(a)$.$x = y \frac{dy}{dx} \cot^2(a) \implies y \frac{dy}{dx} = x 	an^2(a)$.Comparing with $y \frac{dy}{dx} = x \cdot f(a)$, we get $f(a) = 	an^2(a)$.Therefore, $f \left( \frac{2\pi}{3} ight) = 	an^2 \left( \frac{2\pi}{3} ight) = (-\sqrt{3})^2 = 3$.

If $\sec^{-1} \left( \frac{x^2 + y^2}{x^2 - y^2} \right) = 2a$, such that $y \frac{dy}{dx} = x \cdot f(a)$, then the value of $f \left( \frac{2\pi}{3} \right)$ is

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