Question

Verify that each equation is an identity
cscx / (cotx + tanx) = cosx

So I rewrote each trig sign as
(

Answer 1

(1/sinx) / (Cosx/sinx + Sinx/Cosx) = cosx

So now what?

Answer 2

add the fractions in the bottom

Answer 3

I know I should know this, but they don't have the same denominators, so how do I fix that since it is a fraction
in a fraction?

Answer 4

\[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]

Answer 5

or in your case, if you call \(\cos(x)=a\) and \(\sin(x)=b\) it is
\[\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\]

Answer 6

then you have \(a^2+b^2=1\) so the denominator is
\[\frac{1}{ab}\] and
\[\frac{\frac{1}{b}}{\frac{1}{ab}}=\frac{1}{b}\times \frac{ab}{1}=a\]
\

Answer 7

the only one piece of this that had anything to do with trig is that
\[a^2+b^2=\cos^2(x)+\sin^2(x)=1\] the rest is algebra

Answer 8

Thank you so much, you are a grade saver.