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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 $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
Know that: $\tan\theta=\dfrac{1}{\cot\theta}$. Therefore: $$\dfrac{(1+\tan^2\theta)}{(1+\cot^2\theta)}=\dfrac{(1+\tan^2\theta)}{\bigg(1+\dfrac{1}{\tan^2\theta}\bigg)}=\dfrac{(1+\tan^2\theta)}{\bigg(\dfrac{1+\tan^2\theta}{\tan^2\theta}\bigg)}=\tan^2\theta.$$
$$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\csc^2\theta}=\frac{\frac{1}{\cos^2\theta}}{\frac{1}{\sin^2\theta}}=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$
As for your second question that $\tan\theta+\cot\theta=\sec\theta\csc\theta,$ just use the fact that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$ to get $$\tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta}\cdot\frac{1}{\cos\theta}=\sec\theta\cdot\csc\theta$$
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.
Draw a right triangle A and label the sides appropriately as OPP, ADJ, and HYP with angle $\theta$ opposite side OPP, and HYP opposite the right angle. Next generate similar triangle B by multiplying each side of triangle A by $\displaystyle\frac{HYP}{OPP\times ADJ }$.
Triangle B has a hypotenuse $\displaystyle HYP\cdot \frac{HYP}{OPP\times ADJ}=\frac{HYP}{OPP}\cdot\frac{HYP}{ADJ}=\csc\theta\sec\theta$.
Its opposite side is $\displaystyle OPP\cdot\frac{HYP}{OPP\times ADJ}=\frac{HYP}{ADJ}=\sec\theta$.
Its adjacent side is $\displaystyle ADJ\cdot\frac{HYP}{OPP\times ADJ}=\frac{HYP}{OPP}=\csc\theta$.
This triangle is shown in the diagram below.
From the definition of $\tan\theta$ we have
$$\color{green}{\tan\theta}=\frac{opposite}{adjacent}=\color{green}{\frac{\sec\theta}{\csc\theta}}$$
We also see that $$\sin\theta=\frac{opposite}{hypotenuse}=\frac{\sec\theta}{\sec\theta\csc\theta}=\frac{1}{\csc\theta}$$ And $$\cos\theta=\frac{adjacent}{hypotenuse}=\frac{\csc\theta}{\sec\theta\csc\theta}=\frac{1}{\sec\theta}$$ Additionally, using the Pythagorean theorem we have the following identity
$$\color{blue}{\csc^2\theta+\sec^2=\sec^2\theta\csc^2\theta}$$
Now, using these new tools your first problem becomes
$$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\csc^2\theta}=\bigg(\color{green}{\frac{\sec\theta}{\csc\theta}}\bigg)^2=(\color{green}{\tan\theta})^2=\tan^2\theta$$ And your second problem becomes $$\begin{array}{lll} \tan\theta+\cot\theta&=&\frac{\sec\theta}{\csc\theta}+\frac{\csc\theta}{\sec\theta}\\ &=&\frac{\sec^2\theta}{\sec\theta\csc\theta}+\frac{\csc^2\theta}{\sec\theta\csc\theta}\\ &=&\frac{\color{blue}{\sec^2\theta+\csc^2\theta}}{\sec\theta\csc\theta}\\ &=&\frac{\color{blue}{\sec^2\theta\csc^2\theta}}{\sec\theta\csc\theta}\\ &=&\sec\theta\csc\theta \end{array}$$
Note that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$, and $\sin^2\theta+\cos^2\theta=1$. Hence $$ \dfrac{1+\tan^2\theta}{1+\cot^2\theta}=\dfrac{1+\dfrac{\sin^2\theta}{\cos^2\theta}}{1+\dfrac{\cos^2\theta}{\sin^2\theta}}=\left(\dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta+\cos^2\theta}\right)\cdot\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta. $$
$$\large\frac{1+\tan^2\theta}{1+\cot^2\theta}$$
$$\implies\large \frac {\frac{cos^2\theta+sin^2\theta}{cos^2\theta}}{\frac{sin^2\theta+cos^2\theta}{sin^2\theta}}$$
$$\implies \large \frac {\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta}} \\$$
$$\implies \large \frac {sin^2\theta}{cos^2\theta} $$
$$\implies \large \tan^2\theta$$
Here's another geometric proof for the case of acute angles. Construct a right triangle $\triangle ABC$, with base angle $\angle ACB= \theta$ and altitude $AD$ to the hypotenuse $AC$ of unit length.
From the definition of tangent and the similarity of the two right triangles formed by the altitude, we conclude that $|AD|=\tan\theta$ and $|CD|=\cot\theta$. The Pythagorean theorem then implies $|AB|=\sqrt{1+\tan^2\theta}$ and $|BC|=\sqrt{1+\cot^2\theta}$. Taking the tangent of $\theta$ for the whole right triangle then gives $$\tan\theta = \frac{|AB|}{|BC|}=\sqrt{\frac{1+\tan^2\theta}{1+\cot^2\theta}}$$ which squares to the desired identity.
One other way among all these answers would be to remark that $$\tan^2 x(1+\text{cot}^2 x)=\tan^2 x\left(1+\frac{1}{\tan^2 x}\right)$$
You just have to develop the last expression and you'll directly have the result.
$$\frac{1+\tan^2x}{1+\cot^2x}=\frac{1+\tan^2x}{1+\frac1{\tan^2x}}=\tan^2x.$$
