How do you simplify $1/(1+sin x) + 1/(1-sin x)$ ?

Question

31363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-16T19:47:04

Let's say your expression is called $E$ .

First, multiply the first fraction by $"1-sinx"$ and the second by $"1+sinx"$

$E = (1-sinx)/((1+sinx) * (1-sinx)) + (1+sinx)/((1+sinx) * (1-sinx))$

$E = (1 cancel(-sinx) + 1 cancel(+sinx))/((1+sinx) * (1-sinx)) = 2/((1+sinx) * (1-sinx))$

Use the algebraic identity $a^2 - b^2 = (a-b)(a+b)$ . In your case,

$a = 1$ and
$b = sinx$

As a result, the expression that serves as a denominator will become

$(1+sinx) * (1-sinx) = 1^(2) - (sinx)^(2) = 1 -sin^2x$

Therefore, $E$ will be

$E = 2/(1 -sin^2x)$

Remember that $1 - sin^2x = cos^2x$ , so the final form of $E$ will be

$E = color(green)(2/cos^2x)$

How do you simplify 1/(1+sin x) + 1/(1-sin x)1/(1+sin x) + 1/(1-sin x)?