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Question #78a09

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Feb 26, 2017

Well, recall that:

$tan x = \frac{sin x}{cos x}$ $tanx = sinx/cosx$

$cot x = \frac{1}{tan x} = \frac{cos x}{sin x}$ $cotx = 1/tanx = cosx/sinx$

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$

${csc}^{2} x = \frac{1}{{sin}^{2} x}$ $csc^2x = 1/sin^2x$

Thus:

${cot}^{2} x (1 + {tan}^{2} x)$ $cot^2x(1+tan^2x)$

$= \frac{{cos}^{2} x}{{sin}^{2} x} (1 + \frac{{sin}^{2} x}{{cos}^{2} x})$ $= cos^2x/sin^2x(1+sin^2x/cos^2x)$

$= \frac{{cos}^{2} x}{{sin}^{2} x} + 1$ $= cos^2x/sin^2x + 1$

$= \frac{{cos}^{2} x + {sin}^{2} x}{{sin}^{2} x}$ $= (cos^2x + sin^2x)/(sin^2x)$

$= \frac{1}{{sin}^{2} x}$ $= 1/sin^2x$

$= {csc}^{2} x$ $= color(blue)(csc^2x)$

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