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If $\tan (x+y)+\tan (x-y)=1$, find $\frac{d y}{d x}$


We are given with an equation $\tan (x+y)+\tan (x-y)=1$, we have to find $\frac{d y}{d x}$ by using the given equation, so

by differentiating the equation on both sides with respect to $x$, we get,

$\sec ^{2}(x+y)\left[1+\frac{d y}{d x}\right]+\sec ^{2}(x-y)\left[1-\frac{d y}{d x}\right]=0$

$\frac{d y}{d x}\left[\sec ^{2}(x+y)-\sec ^{2}(x-y)\right]+\sec ^{2}(x+y)+\sec ^{2}(x-y)=0$

$\frac{d y}{d x}=\frac{\sec ^{2}(x+y)+\sec ^{2}(x-y)}{\sec ^{2}(x-y)-\sec ^{2}(x+y)}$

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