1. ## Proving Identities problem

The book says

Prove the identity: 1 + cscx/ secx = cosx + cotx

and

1 - 2 secx - 3secx^2x/ -tanx^2 = 1 - 3secx/ 1 - secx

Thanks and explanation and final answer would be great.

2. Prove the identity: $\displaystyle 1 + cscx/ secx = cosx + cotx$

$\displaystyle \cos{x} = 1$ when $\displaystyle x=0$

$\displaystyle \cos{x} + \frac{\frac{1}{\sin{x}}}{\frac{1}{\cos{x}}} = \cos{x} + \cot{x}$

so

$\displaystyle \cos{x} + \cot{x} = \cos{x} + \cot{x}$

on the next one ... is this equation correct as given
before we see if it can be proved

1 - 2 secx - 3secx^2x/ -tanx^2 = 1 - 3secx/ 1 - secx

$\displaystyle 1 - 2 \sec{x} - \frac{3\sec^{2x}{x}}{ -\tan^2{x}} = 1 - \frac{3\sec{x}}{1 - \sec{x}}$

3. this equation is right except for

its supposed to be sec squared of x. I typed it wrong on the initial question my bad.

4. Hello, BHSKid!

Use parentheses please . . . or learn to use LaTeX.

Prove: .(1 + csc x)/(sec x) = cos x + cot x

We have: .$\displaystyle \frac{1 + \csc x}{\sec x} \;=\;\frac{1}{\sec x} + \frac{\csc x}{\sec x} \;=\;\cos x + \frac{\frac{1}{\sin x}}{\frac{1}{\cos x}} \;=\;\cos x + \frac{\cos x}{\sin x} \;=\;\cos x + \cot x$

(1 - 2 sec x - 3 secx^2x)/(-tanx^2) = (1 - 3sec x)/(1 - sec x)

We have: .$\displaystyle \frac{1 - 2\sec x - 3\sec^2\!x}{-\tan^2\!x} \;=\;\frac{(1+\sec x)(1-3\sec x)}{1 - \sec^2\!x} \;=\;\frac{(1+\sec x)(1-3\sec x)}{(1+\sec x)(1-\sec x)}$ .$\displaystyle =\;\frac{1-3\sec x}{1 - \sec x}$