How do you prove $\frac{cot (x) - tan (x)}{sin (x) cos (x)} = {csc}^{2} (x) - {sec}^{2} (x)$ $(Cot(x) - tan(x)) /( sin(x) cos(x)) = csc^2(x) - sec^2(x)$ ?

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Answer 1 · 2016-05-05T20:46:15

see below

Left Side: $= \frac{\frac{cos x}{sin x} - \frac{sin x}{cos x}}{sin x cos x}$ $=(cosx/sinx -sinx/cosx)/(sinxcosx)$

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

$= \frac{{cos}^{2} x - {sin}^{2} x}{sin x cos x} \times \frac{1}{sin x cos x}$ $=(cos^2x-sin^2x)/(sinxcosx) xx 1/(sinxcosx)$

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

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

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

$= {csc}^{2} x - {sec}^{2} x$ $=csc^2x-sec^2x$

$=$ $=$ Right Side

How do you prove (Cot(x) - tan(x)) /( sin(x) cos(x)) = csc^2(x) - sec^2(x)cot(x)−tan(x)sin(x)cos(x)=csc2(x)−sec2(x)(Cot(x) - tan(x)) /( sin(x) cos(x)) = csc^2(x) - sec^2(x)?