Trigonometric Identities help How do you solve this? I can't figure out what I should do. 
$$\sin ^4\left(A\right)+\cos ^2\left(A\right)=\cos ^4\left(A\right)+\sin ^2\left(A\right)$$
Also, why is this equal zero? Can someone explain how that simplifies to be zero? 
$$\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}$$ 
 A: Note that $\sin^4 A-\cos^4 A=(\sin^2 A-\cos^2 A)(\sin^2 A+\cos^2 A)=\sin^2 A-\cos^2 A$.
For the other question, we have $\frac{1}{\cos x}-1=\frac{1-\cos x}{\cos x}$ and $\frac{1}{\cos x}+1=\frac{1+\cos x}{\cos x}$. When we divide, the $\cos x$ cancel, and we get $\frac{1-\cos x}{1+\cos x}$.
A: You just have to know that:
$\sin^4(a)=\sin^2(a)\sin^2(a)$
$\cos^4(a)=\cos^2(a)\cos^2(a)$
$\sin^2(a)+\cos^2(a)=1$

$$\begin{align}
\sin^4(a)+\cos^2(a)&=\cos^4(a)+\sin^2(a) \\
\sin^2(a)\sin^2(a)+\cos^2(a)&=\cos^2(a)\cos^2(a)+\sin^2(a) \\
\sin^2(a)\left(1-\cos^2(a)\right)+\cos^2(a)&=\cos^2(a)\left(1-\sin^2(a)\right)+\sin^2(a) \\
\sin^2(a)-\sin^2(a)\cos^2(a)+\cos^2(a)&=\cos^2(a)-\sin^2(a)\cos^2(a)+\sin^2(a) \\
1-\sin^2(a)\cos^2(a)&=1-\sin^2(a)\cos^2(a) \\
0&=0 \\
\end{align}$$
A: For the 1st problem there's a little trick: add zero.
\begin{array}{lll}\sin^4A+\cos^2A&=&\sin^4A+(-\cos^4A+\cos^4A)+\cos^2A\\
&=&(\sin^4A-\cos^4A)+\cos^4A+\cos^2A\\
&=&(\sin^2A-\cos^2A)(\sin^2A+\cos^2A)+\cos^4A+\cos^2A\\
&=&(\sin^2A-\cos^2A)+\cos^4A+\cos^2A\\
&=&\cos^4A+\sin^2A
\end{array}
For the other problem, the trick is to multiply by one.
\begin{array}{lll}
\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}&=&\frac{\left(\frac{1}{\cos \left(x\right)}\right)-1}{\left(\frac{1}{\cos \left(x\right)}\right)+1}\cdot\frac{\cos(x)}{\cos(x)}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}\\
&=&\frac{1-\cos(x)}{1+\cos(x)}+\frac{\cos \left(x\right)-1}{\cos \left(x\right)+1}\\
&=&\frac{1-\cos(x)}{1+\cos(x)}-\frac{1-\cos(x)}{1+\cos(x)}\\
&=&0
\end{array}
