Trigonometry Problem. Help me! Simplify
 $$\frac{\cos^{2}a-\cot^{2}a +1}{\sin^{2}a + \tan^{2} a -1}$$
Please help me solve this problem
 A: $$\frac{\cos^{2}(a)-\cot^{2}(a)+1}{\sin^{2}(a)+\tan^{2}(a)-1}=\frac{\cos^{2}(a)-\frac{\cos^{2}(a)}{\sin^{2}(a)}+1}{\sin^{2}(a)+\frac{\sin^{2}(a)}{\cos^{2}(a)}-1}$$
$$=\frac{\cos^{2}(a)\left(1-\frac{1}{\sin^{2}(a)}+\frac{1}{\cos^{2}(a)}\right)}{\sin^{2}(a)\left(1+\frac{1}{\cos^{2}(a)}-\frac{1}{\sin^{2}(a)}\right)}=\frac{\cos^{2}(a)\left(1-\csc^{2}(a)+\sec^{2}(a)\right)}{\sin^{2}(a)\left(1+\sec^{2}(a)-\csc^{2}(a)\right)}$$
$$=\frac{\cos^{2}(a)\left(1-\csc^{2}(a)+\sec^{2}(a)\right)}{\sin^{2}(a)\left(1-\csc^{2}(a)+\sec^{2}(a)\right)}=\frac{\cos^{2}(a)}{\sin^{2}(a)}$$
$$=\left(\frac{\cos(a)}{\sin(a)}\right)^{2}=\cot^{2}(a)$$
A: Use


* $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$

* $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

* $\sin^2(\theta)+\cos^2(\theta)=1$

A: The easiest way to attack a tough one like this is to write everything in terms of $s \equiv \sin \alpha$, simplify the resulting compound fraction, and then at the end, see if you recognize some other trig functions in the answer.
Here we have 
$$
\frac{1-s^2-\frac{1-s^2}{s^2}+1}{s^2+\frac{s^s}{1-s^2}-2}
$$
which becomes (multiply numerator and denominator by $s^2(1-s^2)$)
$$
\frac{(3s^2 - s^4 -1)(1-s^2)}{3s^4-s^6-s^2} = \frac{1-s^2}{s^2} 
= \frac{\cos^2 \alpha}{\sin^2 \alpha} = \cot^2 \alpha
$$
