1. trig identities

i understand how to prove them and all
but i just need help proving the following 2:

1) (sinx + tanx)/cosx + 1 = tan x

2) tanx + tan^3x = sec^2x/cot x

thanks

2. Hello, checkmarks!

$\displaystyle 1)\;\;\frac{\sin x + \tan x}{\cos x + 1} \;= \;\tan x$

We have: .$\displaystyle \dfrac{\sin x + \frac{\sin x}{\cos x}}{\cos x + 1}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}$: . $\displaystyle \dfrac{\cos x}{\cos x}\cdot\dfrac{\sin x + \frac{\sin x}{\cos x}}{\cos x + 1} \;=\;\frac{\sin x\cos x + \sin x}{\cos x(\cos x + 1)}$

Factor: .$\displaystyle \frac{\sin x(\cos x + 1)}{\cos x(\cos x + 1)} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x$

$\displaystyle 2)\;\;\tan x + \tan^3\!x \;= \;\frac{\sec^2\!x}{\cot x}$

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

3. thank you so much

i have another one..last one, i promise

#3 (1-cos^2x)(sec^2x - 1) = tan^2x
asgaghkcos^2xagjahajaaha

4. Hello, checkmarks!

This one is not true . . . probably a typo . . .

$\displaystyle 3)\;\;\frac{(1-\cos^2\!x)(\sec^2\!x - 1)}{\cos^2\!x} \:= \:\tan^{{\color{red}4}}\!x$

We have: .$\displaystyle \frac{\overbrace{(1 - \cos^2\!x)}^{\text{This is }\sin^2\!x}\,\overbrace{(\sec^2\!x - 1)}^{\text{This is }\tan^2\!x}}{\cos^2\!x} \;\;=\;\;\frac{\sin^2\!x\cdot\tan^2\!x}{\cos^2\!x} \;=\;\frac{\sin^2\!x}{\cos^2\!x}\cdot\tan^2x$

. . $\displaystyle =\;\;\left(\frac{\sin x}{\cos x}\right)^2\!\!\cdot\tan^2\!x \;\;= \;\;\tan^2\!x\cdot\tan^2\!x \;\;=\;\;\tan^4\!x$