Continuity and Differentiability — Class 12 Maths Solution

Let $u = \frac{x}{{\sin x}}$ and $v = \sin x$
therefore,$\frac{{du}}{{dx}} = \frac{{\sin x \cdot \frac{d}{{dx}}x - x \cdot \frac{d}{{dx}}\sin x}}{{{{(\sin x)}^2}}}$

$= \frac{{\sin x - x\cos x}}{{{{\sin }^2}x}}$ ……(i)

and $\frac{{dv}}{{dx}} = \frac{d}{{dx}}\sin x = \cos x$ …..(ii)

therefore,$\frac{{du}}{{dv}} = \frac{{du/dx}}{{dv/dx}} = \frac{{\sin x - x\cos x/{{\sin }^2}x}}{{\cos x}}$

$= \frac{{\sin x - x\cos x}}{{{{\sin }^2}x\cos x}} = \frac{{\frac{{\sin x - x\cos x}}{{\cos x}}}}{{\frac{{{{\sin }^2}x\cos x}}{{\cos x}}}}$

[Dividing numerator and denominator by cosx ]
$= \frac{{\tan x - x}}{{{{\sin }^2}x}}$