Explanation:
\(\frac{dy}{dx} = \frac{sin(a+y) cosy - siny cos(a+y)}{sin^2(a+y)}\)
\(x = \frac{siny}{sin(a+y)}\)
On differentiating w.r.t. y, we get
\(=\frac{sin(a+y-y)}{sin^2(a+y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{sin^2(a+y)}{sina}\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]