Question

if u=log(tanx+tany+tanz) show that sin2x dou u/dou x+ sin2y dou u/ dou y + sin2z dou u/ dou z=2​

Answer 1

Answer:

Given:

$u=log(tanx+tany+tanz)$

Differentiate partially with respect to x

$\frac{\partial{u}}{\partial{x}}=\frac{1}{tanx+tany+tanz}\:sec^2x$

$\implies\:sin2x\frac{\partial{u}}{\partial{x}}=\frac{sin2x}{tanx+tany+tanz}.\frac{1}{cos^2x}$

$\implies\:sin2x\frac{\partial{u}}{\partial{x}}=\frac{2\:sinx\:cosx}{tanx+tany+tanz}.\frac{1}{cos^2x}$

$\implies\:sin2x\frac{\partial{u}}{\partial{x}}=\frac{2}{tanx+tany+tanz}.\frac{sinx}{cosx}$

$\implies\:sin2x\frac{\partial{u}}{\partial{x}}=\frac{2}{tanx+tany+tanz}.tanx$

$\implies\:sin2x\frac{\partial{u}}{\partial{x}}=\frac{2\:tanx}{tanx+tany+tanz}$

similarly

$sin2y\frac{\partial{u}}{\partial{y}}=\frac{2\:tany}{tanx+tany+tanz}$

$sin2y\frac{\partial{u}}{\partial{z}}=\frac{2\:tanz}{tanx+tany+tanz}$

Adding these, we get

$sin2x\frac{\partial{u}}{\partial{x}}+sin2y\frac{\partial{u}}{\partial{y}}+sin2y\frac{\partial{u}}{\partial{z}}=\frac{2\:tanx}{tanx+tany+tanz}+\frac{2\:tany}{tanx+tany+tanz}+\frac{2\:tanz}{tanx+tany+tanz}$

$sin2x\frac{\partial{u}}{\partial{x}}+sin2y\frac{\partial{u}}{\partial{y}}+sin2y\frac{\partial{u}}{\partial{z}}=\frac{2\:tanx+2\:tany+2\:tanz}{tanx+tany+tanz}$

$sin2x\frac{\partial{u}}{\partial{x}}+sin2y\frac{\partial{u}}{\partial{y}}+sin2y\frac{\partial{u}}{\partial{z}}=\frac{2(tanx+tany+tanz)}{tanx+tany+tanz}$

$\implies\:\boxed{\bf\:sin2x\frac{\partial{u}}{\partial{x}}+sin2y\frac{\partial{u}}{\partial{y}}+sin2y\frac{\partial{u}}{\partial{z}}=2}$