Let $\sin^{-1} \alpha = A$, $\sin^{-1} \beta = B$, $\sin^{-1} \gamma = C$
$A + B + C = \pi$
$(\alpha + \beta)^2 - \gamma^2 = 3 \alpha \beta$
$\alpha^2 + \beta^2 - \gamma^2 = \alpha \beta$
$\frac{\alpha^2 + \beta^2 - \gamma^2}{2 \alpha \beta} = \frac{1}{2}$
$\Rightarrow \cos C = \frac{1}{2}$
$\sin C = \gamma$
$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$
$\gamma = \frac{\sqrt{3}}{2}$
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32