# Thread: Is there a special technique to proving trigonmetric identities?

1. ## Is there a special technique to proving trigonmetric identities?

Like for an example:
Prove:
sin(x) + cos(x)cot(x) = csc(x)

And more complex problems like:
(tan^2(x) / (sec(x) + 1)) = (1-cos(x)) / cos(x)

Is there a special technique to follow?

2. Originally Posted by mwok
Like for an example:
Prove:
sin(x) + cos(x)cot(x) = csc(x)

And more complex problems like:
(tan^2(x) / (sec(x) + 1)) = (1-cos(x)) / cos(x)

Is there a special technique to follow?
In establishing an identity, its best that you work with the side of the equation that seems to be more complex.

In the first one, I'd focus on getting the left side of the equation to look like the right side of the equation:

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

Now, find the LCD, and then combine the fractions:

LCD = $\sin x$

$\therefore\sin x+\frac{\cos^2 x}{\sin x}=\frac{\sin^2x}{\sin x}+\frac{\cos^2x}{\sin x}=\frac{\overbrace{\sin^2x+\cos^2x}^1}{\sin x}=\frac{1}{\sin x}=\color{red}\boxed{\csc x}$

We've established that the LHS = RHS.

Can you try the other one?

I would focus on getting the LHS to look like the RHS...

Does this make sense?

--Chris

3. tan^2(x) / (sec(x) + 1)= (1 - cos(x) )/ cos(x)

I'm stuck at:
sin(x)^2*(1/cos(x)+1)/cos(x)^2

I converted tan^2 into sin/cos (both squared). For the denominator, I converted sec(x) into 1/cos(x) + 1.

What's the next step?

4. Originally Posted by mwok
tan^2(x) / (sec(x) + 1)= (1 - cos(x) )/ cos(x)

I'm stuck at:
sin(x)^2*(1/cos(x)+1)/cos(x)^2

I converted tan^2 into sin/cos (both squared). For the denominator, I converted sec(x) into 1/cos(x) + 1.

What's the next step?
I didn't think of going about it that way...

Hint: Apply the identity $\tan^2x=\sec^2x-1$ and recall that $a^2-b^2=(a+b)(a-b)$

See how far you get now.

--Chris