How do you prove $csc x - \frac{sec x {sin}^{2} x}{tan x} = {cos}^{3} x csc x sec x$ $cscx-(secxsin^2x)/tanx=cos^3xcscxsecx$ ?

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Answer 1 · 2016-05-19T09:42:05

Please see below.

$csc x - \frac{sec x {sin}^{2} x}{tan x}$ $cscx-(secxsin^2x)/tanx$

= $\frac{1}{sin x} - \frac{\frac{1}{cos x} \times {sin}^{2} x}{\frac{sin x}{cos x}}$ $1/sinx-(1/cosx xxsin^2x)/(sinx/cosx)$

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

= $\frac{1}{sin x} - sin x$ $1/sinx-sinx$

= $\frac{1 - {sin}^{2} x}{sin x}$ $(1-sin^2x)/sinx$

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

= ${cos}^{2} x csc x \times cos x \times \frac{1}{cos x}$ $cos^2xcscx xx cosx xx1/cosx$

= ${cos}^{2} x \times csc x \times cos x \times sec x$ $cos^2x xx cscx xx cosx xx secx$

= ${cos}^{3} x csc x sec x$ $cos^3xcscxsecx$

How do you prove cscx-(secxsin^2x)/tanx=cos^3xcscxsecx cscx−secxsin2xtanx=cos3xcscxsecx cscx-(secxsin^2x)/tanx=cos^3xcscxsecx ?