[Math] How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$

Question

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 $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$

Answer 1

Know that: $\tan\theta=\dfrac{1}{\cot\theta}$. Therefore: $$\dfrac{(1+\tan^2\theta)}{(1+\cot^2\theta)}=\dfrac{(1+\tan^2\theta)}{\bigg(1+\dfrac{1}{\tan^2\theta}\bigg)}=\dfrac{(1+\tan^2\theta)}{\bigg(\dfrac{1+\tan^2\theta}{\tan^2\theta}\bigg)}=\tan^2\theta.$$