Question #0c317

Question

1635 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-13T23:52:14

$\frac{1}{sin (x) cos (x)}$ $1/ (sin(x)cos(x))$

Remember the trigonometric identities

So

$cot (x) {sec}^{2} (x)$ $cot(x)sec^2(x)$

$= cot (x) [{tan}^{2} (x) + 1]$ $=cot(x)[tan^2(x) + 1]$

$= tan (x) + cot (x)$ $=tan(x) + cot(x)$

$= (\frac{sin (x)}{cos (x)}) + (\frac{cos (x)}{sin (x)})$ $=(sin(x)/cos(x)) + (cos(x)/sin(x))$

$= (\frac{{sin}^{2} (x)}{sin (x) cos (x)}) + (\frac{{cos}^{2} (x)}{sin (x) cos (x)})$ $=(sin^2(x)/(sin(x)cos(x))) + (cos^2(x)/(sin(x)cos(x)))$

$= \frac{{sin}^{2} (x) + {cos}^{2} (x)}{sin (x) cos (x)}$ $=(sin^2(x) + cos^2(x))/ (sin(x)cos(x))$

$= \frac{1}{sin (x) cos (x)}$ $=1/(sin(x)cos(x))$