## Algebra and Trigonometry 10th Edition

The identity is verified. $\sqrt {\frac{1+sin~\theta}{1-sin~\theta}}=\frac{1+sin~\theta}{|cos~\theta|}$
$\sqrt {\frac{1+sin~\theta}{1-sin~\theta}}=\sqrt {\frac{1+sin~\theta}{1-sin~\theta}}\sqrt {\frac{1+sin~\theta}{1+sin~\theta}}=\sqrt {\frac{(1+sin~\theta)^2}{1-sin^2\theta}}=\sqrt {\frac{(1+sin~\theta)^2}{cos^2\theta}}=\frac{|1+sin~\theta|}{|cos~\theta|}$ But, since $1+sin~\theta\geq0$ for any value of $\theta$: $|1+sin~\theta|=1+sin~\theta$ $\sqrt {\frac{1+sin~\theta}{1-sin~\theta}}=\frac{1+sin~\theta}{|cos~\theta|}$