# Thread: Proving a trig idenity.

1. ## Proving a trig idenity.

I've tried a few different things for this one, but I can't seem to figure it out.

Could you point in the right direction and tell me what identity I should try to use to figure this one out.

$cotx (tanx+cotx)=\frac{1}{sin^2x}$

2. Originally Posted by mtt0216
I've tried a few different things for this one, but I can't seem to figure it out.

Could you point in the right direction and tell me what identity I should try to use to figure this one out.

$\cot x (\tan x+\cot x)=\frac{1}{\sin^2x}$
$\cot(x) \tan(x) + \cot^2(x) = 1+ \cot ^2(x) = 1+\frac{\cos^2(x)}{\sin^2(x)}$

Rewrite 1 as $\frac{\sin^2(x)}{\sin^2(x)}$.

This gives $\frac{\sin^2(x) + \cos^2(x)}{\sin^2(x)}$

I'm sure you know what $\sin^2(x) + \cos^2(x)$ is equal to

3. Originally Posted by mtt0216
I've tried a few different things for this one, but I can't seem to figure it out.

Could you point in the right direction and tell me what identity I should try to use to figure this one out.

$cotx(tanx+cotx)=\frac{1}{sin^2x}$
Just start by writing out the LHS using Sinx, since that is what we have on the RHS

$cotx(tanx+cotx)=\frac{cosx}{sinx}\left(\frac{sinx} {cosx}+\frac{cosx}{sinx}\right)$

$=\frac{cosx}{sinx}\ \frac{sinx}{cosx}+\frac{cosx}{sinx}\ \frac{cosx}{sinx}$

$=1+\frac{cos^2x}{sin^2x}$

Now finish up by writing 1 with $sin^2x$ in the denominator

$1=\frac{sin^2x}{sin^2x}$

As they have a common denominator, you can combine the numerators

and use the important identity $sin^2x+cos^2x=1$

to complete

4. Well thanks for the help, but I did it a different way that seemed easier to me. Can you check to see if it is correct.

$cotx(tanx+cotx)=\frac{1}{sin^2x}$

$
cotx tanx +cot^2x$

$
\frac{1}{tanx} \frac{tanx}{1} =1$

$1 + cot^2x$

$=csc^2x$

= $\frac{1}{sinx} \frac{1}{sinx}$

= $\frac{1}{sin^2x}$

5. Yes mtt0216,

though it would be more adviseable for you to be able to show why

$1+cot^2x=csc^2x$

as e^(i*pi) showed.

It's good practice at this stage for you,
and helps you find the calculations getting simpler.
As you say, you found that way easier because it allowed you
to bypass that step using another "less obvious" identity.