u$ = \; log \;( \; tan x \; + \; tan y \; ) \\ \therefore \; \frac{\partial u}{\partial x} \; = \frac{\partial }{\partial x} [ log(tan \; x \; + \; tan \; y \; )] \; = \; \frac{1}{tan \; x \; + \; tan \; y } \; \frac{\partial }{\partial x} (tan x \; + \; tan y \; ) \\ \; \\ = \frac{sec^2\;x \; + \; 0}{tan x \; + \; tan y} \; = \frac{sec^2\;x }{tan x \; + \; tan y} \;\;\;\;\;\;\;\;\;\;\;\; \ldots (i) \\ \; \\ Similarly, \; \; \frac{\partial u}{\partial y} \; = \frac{\partial }{\partial y} [ log(tan \; x \; + \; tan \; y \; )] \; = \; \frac{sec^2\;y }{tan x \; + \; tan y} \;\;\;\;\;\;\;\;\;\;\;\; \ldots (ii) \\ \; \\ \therefore L.H.S. \; = \; sin \; 2x \; \frac{\partial u}{\partial x} \; + \; + sin \;2y \; \frac{\partial u}{\partial y} \\ \; \\ = 2 \; sinx \cdot cosx \bigg( \frac{sec^2\;x }{tan x \; + \; tan y} \bigg) \; \; + \; \; 2 \; siny \cdot cosy \bigg( \frac{sec^2\;y }{tan x \; + \; tan y} \bigg) \\ \; \\ = \frac{2(sin\;x/cosx)}{tan \;x \; + \; tan \;y} \; \; + \; \; \frac{2(sin \;y/cosy)}{tan \;x \; + \; tan \;y} \{ \because \; sec^2 \; x \; = \; \frac{1}{cos^2x}\\ \; \; \; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sec^2 \; y \; = \; \frac{1}{cos^2 \; y} \} \\ \; \\ = \frac{2 \; tanx \; + \; 2 \; tany}{tanx \; + \; tany} \; = \; 2 \; = \; RHS \ \ \ldots \;Hence \; \; Proved $