How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$ 
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$

Look at $\sin^2 \theta + \cos^2 \theta = 1$. Dividing by $\cos^2 \theta$ we get $\tan^2 \theta + 1 = \sec^2 \theta$. Try to find another relation by dividing by $\sin^2 \theta $ and see what appears. 
