# How do you verify  (1+sinx)/(1-sinx) - (1-sinx)/(1+sinx)=4tanxsecx?

Nov 1, 2015

See explanation.

#### Explanation:

$\left[1\right] \text{ } \frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x}$

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

$\left[2\right] \text{ } = \left(\frac{1 + \sin x}{1 - \sin x}\right) \left(\frac{1 + \sin x}{1 + \sin x}\right) - \left(\frac{1 - \sin x}{1 + \sin x}\right) \left(\frac{1 - \sin x}{1 - \sin x}\right)$

$\left[3\right] \text{ } = \frac{1 + 2 \sin x + {\sin}^{2} x}{1 - {\sin}^{2} x} - \frac{1 - 2 \sin x + {\sin}^{2} x}{1 - {\sin}^{2} x}$

$\left[4\right] \text{ } = \frac{1 + 2 \sin x + {\sin}^{2} x - 1 + 2 \sin x - {\sin}^{2} x}{1 - {\sin}^{2} x}$

$\left[5\right] \text{ } = \frac{4 \sin x}{1 - {\sin}^{2} x}$

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

$\left[6\right] \text{ } = \frac{4 \sin x}{{\cos}^{2} x}$

$\left[7\right] \text{ } = \frac{4 \sin x}{\left(\cos x\right) \left(\cos x\right)}$

Quotient Identity: $\sin \frac{\theta}{\cos} \theta = \tan \theta$

$\left[8\right] \text{ } = \frac{4 \tan x}{\cos x}$

Reciprocal Identity: $\frac{1}{\cos} \theta = \sec \theta$

$\left[9\right] \text{ } = 4 \tan x \sec x$

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