Question

I need help proving a trig identity, (sec^2(x)csc(x))/(sec^2(x)+csc^2(x))

It's been over a year since I've taken math and I'm about to rip my hair out over this question.

Answer 1

Forgot to add, it equals sin(x)

Answer 2

sec^2x * cscx = 1/ cos^2x * 1/ sinx
sec^2x + csc^2x = 1/cos^2x + 1/sin^2x = 1 / cos^2xsin^2x
dividing first by second gives sin x

Answer 3

get it ?

Answer 4

Using what jimmy said in latex it would read:
\[\frac{\sec^2(x)\csc(x)}{\sec^2(x)+\csc^2(x)}=\sin(x)\]
Use the manipulation that:\[\sec^2(x)\csc(x)=\frac{1}{\cos^2(x)}*\frac{1}{\sin(x)}\]
And:
\[\sec^2(x)+\csc^2(x)=\frac{1}{\cos^2(x)}+\frac{1}{\sin^2(x)}=\frac{1}{\cos^2(x)\sin^2(x)}\]
So:
\[\frac{\frac{1}{\cos^2(x)\sin(x)}}{\frac{1}{\cos^2(x)\sin^2(x)}}=\frac{\cos^2(x)\sin^2(x)}{\cos^2(x)\sin(x)}=\sin(x)\]
After you divide it out.