1. [SOLVED] Proving Identities

Hi,
I am realllly bad at proving identities... so I would appreciate any advice you have, when proving these three identities:

(4/4a+4) + (7/7a-7) = (2a^2 + 2a/a^2 + 2a + 1) * (a+2)/(a^2+a-2)

also:

tan^2x/sec^2x = sin^2x

also:

(1+tan^2x)(a-sin^2x)=1

thank you for any guidance you have to offer

2. for the 2nd one, so far i have:

tan^2x/sec^2x = sin^2x

[(sinx/cosx)^2]/ (1/cos^2x) = sin^2x

that's all, lol

3. Hello, ria3!

The first one is ugly.
There is no neat way to transform one side to equal the other.

$\displaystyle \frac{4}{4a+4} + \frac{7}{7a-7} \;=\; \frac{2a^2 + 2a}{a^2 + 2a + 1}\cdot\frac{a+2}{a^2+a-2}$

$\displaystyle \frac{\tan^2\!x}{\sec^2\!x} \:=\:\sin^2\!x$
The left side is: .$\displaystyle \frac{\dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos ^2\!x}} \;=\;\frac{\sin^2\!x}{\cos^2\!x}\cdot\frac{\cos^2\ !x}{1} \;=\;\sin^2\!x$

$\displaystyle (1+\tan^2\!x)(1-\sin^2\!x)\:=\:1$
This one is too easy!

We have: .$\displaystyle \underbrace{(1 + \tan^2\!x)}_{\text{This is }\sec^2\!x}\underbrace{(1 -\sin^2\!x)}_{\text{This is }\cos^2\!x} \;=\;\sec^2\!x\cdot\cos^2\!x \;=\;1$

4. thank you soraban! i really appreciate it.
i think i've figured out the 1st. it turns out all you have to do is simplify each side by factoring. so, both sides end up being 2a/(a+1)(a-1)