How do I simplify $\frac{1 + {tan}^{2} x + {sec}^{2} x {cot}^{2} x}{{csc}^{2} x + {cot}^{2} x {csc}^{2} x}$ $(1+tan^2x+sec^2xcot^2x)/(csc^2x+cot^2xcsc^2x)$ ?

Question

$\frac{1 + {tan}^{2} x + {sec}^{2} x {cot}^{2} x}{{csc}^{2} x + {cot}^{2} x {csc}^{2} x}$ $(1+tan^2x+sec^2xcot^2x) / (csc^2x+cot^2xcsc^2x)$

I tried simplifying terms and it turned into a mega mess. I just need a simple explanation. Thank you!

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Answer 1 · 2018-02-06T01:10:06

$\frac{1 + {tan}^{2} x + {sec}^{2} x {cot}^{2} x}{{csc}^{2} x + {cot}^{2} x {csc}^{2} x}$ $(1+tan^2x+sec^2xcot^2x) / (csc^2x+cot^2xcsc^2x)$

$= \frac{{sec}^{2} x + {sec}^{2} x {cot}^{2} x}{{csc}^{2} x + {cot}^{2} x {csc}^{2} x}$ $=(sec^2x+sec^2xcot^2x) / (csc^2x+cot^2xcsc^2x)$

$=(sec^2x(cancel(1+cot^2x)) )/ (csc^2x(cancel(1+cot^2x))$

$=(1/cos^2x)/(1/sin^2x)$

$=sin^2x/cos^2x=tan^2x$