How do you simplify the expression $\frac{tan θ + cot θ}{{csc}^{2} θ}$ $(tantheta+cottheta)/csc^2theta$ ?

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Answer 1 · 2016-09-19T13:47:13

$tan θ$ $tantheta$ .

We will use the Trigo. Identity $: 1 + {cot}^{2} x = {csc}^{2} x$ $: 1+cot^2x=csc^2x$

Now, the Exp. $= \frac{tan θ + cot θ}{{csc}^{2} θ}$ $=(tantheta+cottheta)/csc^2theta$

$= \frac{\frac{1}{cot θ} + cot θ}{{csc}^{2} θ}$ $=(1/cottheta +cottheta)/csc^2theta$

$= \frac{\frac{1 + {cot}^{2} θ}{cot θ}}{{csc}^{2} θ}$ $=((1+cot^2theta)/cottheta)/csc^2theta$

$= \frac{\frac{{csc}^{2} θ}{cot θ}}{{csc}^{2} θ}$ $=(csc^2theta/cottheta)/csc^2theta$

$= \frac{1}{cot θ}$ $=1/cottheta$

$= tan θ$ $=tantheta$ .

How do you simplify the expression tanθ+cotθcsc2θ(tantheta+cottheta)/csc^2theta?