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$\frac{1 + sec x}{{tan}^{2} x} = \frac{cos x}{1 - cos x}$ $(1+secx)/(tan^(2)x)=(cosx)/(1-cosx)$

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Answer 1 · 2018-03-03T05:03:20

$= L . H . S$ $=L.H.S$

$= \frac{1 + sec x}{{tan}^{2} x}$ $=(1+secx)/(tan^2x)$

$= ⎛ ⎝ \frac{1 + \frac{1}{cos x}}{\frac{{sin}^{2} x}{{cos}^{2} x}} ⎞ ⎠$ $=((1+1/cosx)/(sin^2x/cos^2x))$

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

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

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

$=(cancelcolor(blue)((cosx+1))cosx)/(cancelcolor(blue)((1+cosx))(1-cosx))$

$=cosx/(1-cosx)$

$=R.H.Scolor(green)([Proved.])$

Answer 2 · 2018-03-03T05:07:32

See below

$(1+secx)/tan^2x=cosx/(1-cosx)$

$(1+secx)/(1-sec^2x)=cosx/(1-cosx)$

$(1+secx)/((1+secx)(1-secx))=cosx/(1-cosx)$

$1/(1-secx)=cosx/(1-cosx)$

$1/(cosx/cosx-1/cosx)=cosx/(1-cosx)$

$1/((cosx-1)/cosx)=cosx/(1-cosx)$

$cosx/(1-cosx)=cosx/(1-cosx)$