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I feel like this problem is simple, but I am having a hard time wrapping my head around it.

$$\dfrac{1}{sin^2(x)}+ \dfrac{sec^2(x)}{tan^2(x)}$$

Any suggestions?

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1 Answer 1

up vote 2 down vote accepted

$$\begin{align}\dfrac{1}{\sin^2(x)}+\dfrac{\sec^2(x)}{\tan^2(x)}&=\dfrac{1}{\sin^2(x)}+\dfrac{\dfrac{1}{\cos^2(x)}}{\dfrac{\sin^2(x)}{\cos^2(x)}}\\ &=\dfrac{1}{\sin^2(x)}+\dfrac{1}{\sin^2(x)}\\ &=\dfrac{2}{\sin^2(x)}\end{align}$$ You can keep going like this. So some more equalities are: $$\dfrac{2}{\sin^2(x)}=2\csc^2(x)=2+2\cot^2(x)$$

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Thanks, the book actually had the answer as $2csc^2(x)$, which is what I think you meant on the bottom there, instead of secant. –  Tanner Feb 27 at 4:05
    
You are correct. I mistyped, I will fix it now. –  jnh Feb 27 at 4:06
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