Let f(x) = cosec x cot x ⇒ f (x) = $\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}$ ⇒ f (x) = $\frac{\cos x}{\sin^2 x}$ ⇒ f (x) = $\cos x (\sin x)^{-2}$ ... (i) Differentiating (i) with respect to x, we get $\frac{d}{dx}$ [f (x)] = (- sin x) $(\sin x)^{-2}$ + (- 2) $(\sin x)^{-3}$ . cos x cos x = – $(\sin x)^{-1}$ – 2 $(\sin x)^{-3}$·cos2x = - $(\sin x)^{-1} - \frac{2\cos^2 x}{\sin^3 x}$ = - $(\sin x)^{-1}$ - $2\cot^2 x$ cosec x = - cosex c - 2 $\cot^2$ x cosec x = – cosec x [1 + 2 $\cot^2$ x] = – cosec x [1 + $\cot^2 x + \cot^2 x$] ∴ $\frac{d}{dx}$ (cosec x cot x) = - cosec x $[\csc^2 x + \cot^2 x]$ = - $\csc^3 x - \csc x \cdot \cot^2 x$