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If $z=\log(\tan x+\tan y) $ then prove that $\sin 2\dfrac {dz}{dx}+\sin 2y \dfrac {dz}{dy}=2.$
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written 7.2 years ago by | • modified 7.2 years ago |
$$z = \log (\tan x+\tan y )…………………………given$$
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 {\sin2z}{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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