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If $z=\log(\tan x+\tan y) $ then prove that $\sin 2\dfrac {dz}{dx}+\sin 2y \dfrac {dz}{dy}=2.$
partial differentiation • 571  views
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$$z = \log⁡ (\tan⁡ x+\tan⁡ y )…………………………given$$

Taking antilog on both sides we get

$e^z=\tan x+\tan ⁡y-------------(1) $

Differentiating w.r.t. x

We get,

$e^z = \dfrac {∂z}{∂x} =\sec^2 x+0 \\ ∴\dfrac {∂z}{∂x} = \dfrac 1{e^z \cos^2 x}$

Multiplying $\sin 2x$ on both side we get

$\sin 2x \dfrac {∂z}{∂x} = \dfrac {\sin⁡2z}{e^z \cos^2 x} \\ =\dfrac {2 \sin x \cos⁡ x}{e^z \cos^2 x} \\ ∴sin 2x \dfrac {∂z}{∂x} =\dfrac {2 \tan ⁡x}{e^z} ---------(2) $

Similar we can have

$\sin 2y \dfrac {∂z}{∂y} =\dfrac {2 \tan ⁡y}{e^z} ----------(3) $

∴Adding eq.(2) & (3) we get

$\sin⁡ 2x \dfrac {∂z}{∂x} + \sin2y \dfrac {∂z}{∂y} = \dfrac {2 \tan ⁡x}{e^z} + \dfrac {2 \tan ⁡y}{e^z}\\ =2e^{-z} (\tan ⁡x + \tan ⁡y ) \\ =2e^{-z}.e^z\space\space from (1) \\ =z $

Hence Proved.

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