Question

How do I get the answer

Answer 1

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ sec(\theta)=\cfrac{1}{cos(\theta)}\\\\ -------------------------------\\\\ \cfrac{1}{1+sin(\theta)}+\cfrac{1}{1-sin(\theta)}\impliedby LCD=[1-sin(\theta)][1+sin(\theta)] \\\\\\ \cfrac{1\underline{-sin(\theta)}+1\underline{+sin(\theta)}}{[1-sin(\theta)][1+sin(\theta)]}\implies \cfrac{2}{[1-sin(\theta)][1+sin(\theta)]}\\\\ -------------------------------$

$\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -------------------------------\\\\ \cfrac{2}{1^2-sin^2(\theta)}\implies \cfrac{2}{1-sin^2(\theta)}\implies \cfrac{2}{cos^2(\theta)} \\\\\\ \cfrac{2}{1}\cdot \cfrac{1^2}{cos^2(\theta)}\implies 2sec^2(\theta)$

Answer 2

Start by finding a common denominator:

1/(1 + sin x) + 1/(1 - sin x)

(1 - sin x)/(1 - sin^2x) + (1 + sin x)/(1 - sin^2x)

Simplify the top and use Pythagorean identity to simplify the bottom:

2/cos^2x

Use the reciprocal identity to simplify the expression:

2sec^2x