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If $\tan (x+y)+\tan (x-y)=1$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x-y)-\sec ^2(x+y)}$
$\tan (x+y)+\tan (x-y)=1$
Differentiating w.r.t. $x$ of $y$, we get
$\begin{aligned} & \Rightarrow \sec ^2(x+y)\left(1+\frac{d y}{d x}\right)+\sec ^2(x-y)\left(1-\frac{d y}{d x}\right)=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x-y)-\sec ^2(x+y)}\end{aligned}$
Differentiating w.r.t. $x$ of $y$, we get
$\begin{aligned} & \Rightarrow \sec ^2(x+y)\left(1+\frac{d y}{d x}\right)+\sec ^2(x-y)\left(1-\frac{d y}{d x}\right)=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{\sec ^2(x+y)+\sec ^2(x-y)}{\sec ^2(x-y)-\sec ^2(x+y)}\end{aligned}$
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