How do I solve #(cot(x)+tan(x)) / (csc(-x))#?

Question

I assume I need to convert #cot(x) + tan(x)# into terms of cosine and sine, then end up with #1/(sin(x)cos(x))#, but I get stuck with how to deal with the rest of the problem from there.

7355 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-14T15:55:07

This equals #-secx#.

I like to rewrite in terms of sine and cosine.

#=(cosx/sinx + sinx/cosx)/(1/sin(-x))#

We also know that #sin(-x) = -sin(x)#.

#= ((cos^2x+ sin^2x)/(cosxsinx))/(-1/sinx)#

We can use #sin^2x + cos^2x = 1#, as you have done.

#= (1/(cosxsinx))/(-1/sinx)#

#= 1/(cosxsinx) * -sinx#

#= -1/cosx#

This can be rewritten using #secx = 1/cosx#.

#=-secx#

Hopefully this helps!