Question #da28a

Question

988 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-07T01:25:36

See below.

We know that $1 - {sin}^{2} x = {cos}^{2} x$ $1-sin^2x=cos^2x$ . This comes handy later.

$\frac{8}{1 - sin x} + \frac{8}{1 + sin x}$ $8/(1-sinx)+8/(1+sinx)$

We can take out an $8$ $8$ from both fractions.
Multiply by the common denominators, and then add the two fractions.

$\frac{8}{1 - sin x} (1 + sin x) + \frac{8}{1 + sin x} (1 - sin x)$ $8/(1-sinx)(1+sinx)+\frac(8)(1+sinx)(1-sinx)$

$8 (\frac{1 + sin x}{1 - {sin}^{2}} + \frac{1 - sin x}{1 - {sin}^{2}})$ $8(\frac(1+sinx)(1-sin^2)+\frac(1-sinx)(1-sin^2))$

$= 8 (\frac{2}{{cos}^{2}})$ $=8(\frac(2)(cos^2))$

$= \frac{16}{{cos}^{2} x}$ $=16/cos^2x$

$= 16 {sec}^{2} x$ $=16sec^2x$