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  • <img alt="\mathrm{\text { If } \tan (x+y)=e^{x+y} \text {, then } \frac{d y}{d x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%20%7B%20If%20%7D%20%5Ctan%20%28x&plu

\mathrm{\text { If } \tan (x+y)=e^{x+y} \text {, then } \frac{d y}{d x}}

Option: 1

is always equal to -1
 


Option: 2

may or may not be equal to -1
 


Option: 3

\frac{\mathrm{dy}}{\mathrm{dx}} cannot be obtained
 


Option: 4

none of these .


Answers (1)

best_answer

\mathrm{\text { Let } x+y=z \text {, Then, } \tan z=e^z}

The equation is satisfied for some particular values of z say \mathrm{z_1, z_2, z_3, \ldots.}

where \mathrm{z_1, z_2, z_3 \ldots} are constants.

\mathrm{\Rightarrow z=x+y=c \, \, is\, \, the \, \, solution \, \, where\, \, c \in\left\{z_1, z_2, z_3, \ldots\right\}}
Differentiating, \mathrm{\frac{d y}{d x}=-1}

Alternate

\mathrm{ \begin{aligned} & \tan (x+y)=e^{x+y}\left[1+\frac{d y}{d x}\right]=e^{x+y}\left[1+\frac{d y}{d x}\right] \\ & \Rightarrow \sec ^2(x+y) \\ & \therefore \frac{d y}{d x}=-1 \text { or } 1+e^{2(x+y)}=e^{x+y} \quad \text { not possible) } \\ & 1+t^2-t=0 \Rightarrow\left(t-\frac{1}{t}\right)^2+\frac{3}{4}=0 \\ & \therefore \frac{d y}{d x}=-1 \quad \forall x, y . \end{aligned} }

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