Proving Identities. I tried to solve it but I cant get the answer. How to prove this by using a hand?


*

*$$ \sec^2x + \csc^2x = \sec^2x \csc^2x $$

*$$ \frac{\sec\theta + 1}{\sec\theta - 1} = \frac{1 + \cos\theta}{1 - \cos\theta}$$

*$$ \frac{1 - \cot^2\theta}{1 + \cot^2\theta} = \sin^2\theta - \cos^2\theta $$


Anyone can help? Thanks.
 A: Hints


*

*$(1)$  Use the fact that $\sin^{2}(x) + \cos^{2}(x) =1$.

*$(2)$ $\displaystyle\sec\theta = \frac{1}{\cos\theta}$

*$(3)$ $\displaystyle\cot\theta = \frac{\cos\theta}{\sin\theta}$
A: Hints:


*

*Write each of $\sec^2$ and $\csc^2$ in terms of $\cos$ and $\sin$, then simplify your fraction using a known trigonometric identity, try to recognize an expression for $\sec^2$ and $\csc^2$ after you've done those steps.

*Try to write $\sec$ in terms of $\cos$ then simplify things and see how far you can get.

*Same thing as (3), write $\cot$ in terms of $\cos$ and $\sin$ and simplify things.

A: I think that #1 is an unfair question, because there is nothing to prove. For if we can assume that
$$\cos^2\theta+ \sin^2\theta=1$$
$$\tan^2\theta+ 1=\sec^2\theta$$
$$\text{and }\cot^2\theta+ 1=\csc^2\theta$$
Then it should be natural to assume that
$$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$
Consider a right triangle with an acute angle $\theta$. Let the hypotenuse be of length $c$, the side adjacent to $\theta$ be of length $a$, and the side opposite angle $\theta$ be of length b.
By the Pythagorean theorem  we have
$$\begin{array}{ll}a^2+b^2=c^2&(1)\end{array}$$
Dividing (1) by $c^2$ we have
$$\cos^2\theta+ \sin^2\theta=1$$
Dividing (1) by $a^2$ we have
$$1+\tan^2\theta = \sec^2\theta$$
Dividing (1) by $b^2$ we have
$$\cot^2\theta + 1= \csc^2\theta$$
Multiplying (1) by $\frac{c^2}{a^2b^2}$ we have
$$\frac{c^2}{b^2}+\frac{c^2}{a^2}=\frac{c^2}{a^2}\cdot\frac{c^2}{b^2}$$
$$\csc^2\theta+\sec^2\theta=\sec^2\theta\csc^2\theta$$
