If (x+y)+ (x+y)=1, find dy dx - Vidyadip

Question

If $\tan(\text{x}+\text{y})+\tan(\text{x}+\text{y})=1,$ find $\frac{\text{dy}}{\text{dx}}$

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Answer

We have, $\tan(\text{x}+\text{y})+\tan(\text{x}-\text{y})=1$
Differentiating with respect to x, we get,
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\tan(\text{x}+\text{y})+\frac{\text{d}}{\text{dx}}\tan(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})=0$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]+\sec^2(\text{x}-\text{y})\Big[1-\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\ \sec^2(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\sec^2(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}} \\ =-\big[\sec^2(\text{x}-\text{y})+\sec^2(\text{x}-\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\sec^2(\text{x}+\text{y})-\sec^2(\text{x}-\text{y})\big] \\ =-\big[\sec^2(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2(\text{x}+\text{y})+\sec^2(\text{x}-\text{y})}{\sec^2(\text{x}-\text{y})-\sec^2(\text{x}+\text{y})}$

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