How do you verify $\frac{1 + sin x}{1 - sin x} - \frac{1 - sin x}{1 + sin x} = 4 tan x sec x$ $(1+sinx)/(1-sinx) - (1-sinx)/(1+sinx)=4tanxsecx$ ?

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How do you verify $\frac{1 + sin x}{1 - sin x} - \frac{1 - sin x}{1 + sin x} = 4 tan x sec x$ $(1+sinx)/(1-sinx) - (1-sinx)/(1+sinx)=4tanxsecx$ ?

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Answer 1 · 2015-11-01T04:47:37

See explanation.

Explanation:

$[1] \frac{1 + sin x}{1 - sin x} - \frac{1 - sin x}{1 + sin x}$ $[1]" "(1+sinx)/(1-sinx)-(1-sinx)/(1+sinx)$

Combine the two terms by making them have the same denominator.

$[2] = (\frac{1 + sin x}{1 - sin x}) (\frac{1 + sin x}{1 + sin x}) - (\frac{1 - sin x}{1 + sin x}) (\frac{1 - sin x}{1 - sin x})$ $[2]" "=((1+sinx)/(1-sinx))((1+sinx)/(1+sinx))-((1-sinx)/(1+sinx))((1-sinx)/(1-sinx))$

$[3] = \frac{1 + 2 sin x + {sin}^{2} x}{1 - {sin}^{2} x} - \frac{1 - 2 sin x + {sin}^{2} x}{1 - {sin}^{2} x}$ $[3]" "=(1+2sinx+sin^2x)/(1-sin^2x)-(1-2sinx+sin^2x)/(1-sin^2x)$

$[4] = \frac{1 + 2 sin x + {sin}^{2} x - 1 + 2 sin x - {sin}^{2} x}{1 - {sin}^{2} x}$ $[4]" "=(1+2sinx+sin^2x-1+2sinx-sin^2x)/(1-sin^2x)$

$[5] = \frac{4 sin x}{1 - {sin}^{2} x}$ $[5]" "=(4sinx)/(1-sin^2x)$

Pythagorean Identity: $1 - {sin}^{2} θ = {cos}^{2} θ$ $1-sin^2theta=cos^2theta$

$[6] = \frac{4 sin x}{{cos}^{2} x}$ $[6]" "=(4sinx)/(cos^2x)$

$[7] = \frac{4 sin x}{(cos x) (cos x)}$ $[7]" "=(4sinx)/((cosx)(cosx))$

Quotient Identity: $\frac{sin θ}{cos θ} = tan θ$ $sintheta/costheta=tantheta$

$[8] = \frac{4 tan x}{cos x}$ $[8]" "=(4tanx)/(cosx)$

Reciprocal Identity: $\frac{1}{cos θ} = sec θ$ $1/costheta=sectheta$

$[9] = 4 tan x sec x$ $[9]" "=4tanxsecx$

$color(blue)("":.(1+sinx)/(1-sinx)-(1-sinx)/(1+sinx)=4tanxsecx)$

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How do you verify $\frac{1 + sin x}{1 - sin x} - \frac{1 - sin x}{1 + sin x} = 4 tan x sec x$ $(1+sinx)/(1-sinx) - (1-sinx)/(1+sinx)=4tanxsecx$ ?

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Explanation:

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How do you verify $\frac{1 + sin x}{1 - sin x} - \frac{1 - sin x}{1 + sin x} = 4 tan x sec x$ $(1+sinx)/(1-sinx) - (1-sinx)/(1+sinx)=4tanxsecx$ ?