How do you verify $\frac{{cot}^{2} x}{1 + csc x} = \frac{1 - sin x}{sin x}$ $cot^2 x/ (1+csc x) = (1-sin x)/ sin x$ ?

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How do you verify $\frac{{cot}^{2} x}{1 + csc x} = \frac{1 - sin x}{sin x}$ $cot^2 x/ (1+csc x) = (1-sin x)/ sin x$ ?

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Answer 1 · 2015-08-12T13:07:03

${sin}^{2} A + {cos}^{2} A = 1$ $sin^2 A + cos^2 A = 1$
Divide by ${sin}^{2} A$ $sin^2 A$
$1 + {cot}^{2} A = {csc}^{2} A$ $1 + cot^2 A = csc^2 A$

$\frac{{cot}^{2} x}{1 + csc x} = \frac{{csc}^{2} x - 1}{1 + csc x} = \frac{(csc x - 1) (csc x + 1)}{1 + csc x} = \frac{1}{sin x} - 1 = \frac{1 - sin x}{sin x}$ $cot^2 x/(1+csc x) = (csc^2 x-1)/(1+csc x) = ((csc x-1)(csc x+1))/(1+csc x) = 1/sin x - 1 = (1 - sin x)/sin x$

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How do you verify $\frac{{cot}^{2} x}{1 + csc x} = \frac{1 - sin x}{sin x}$ $cot^2 x/ (1+csc x) = (1-sin x)/ sin x$ ?

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How do you verify cot^2 x/ (1+csc x) = (1-sin x)/ sin xcot2x1+cscx=1−sinxsinxcot^2 x/ (1+csc x) = (1-sin x)/ sin x?

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How do you verify $\frac{{cot}^{2} x}{1 + csc x} = \frac{1 - sin x}{sin x}$ $cot^2 x/ (1+csc x) = (1-sin x)/ sin x$ ?