i have a trig proof that i started but ended up going no where with i got like 4 steps into it and hit a wall. if someone could help me out it would be much appriciated thanx. Sec+2tan = cot+2cos
>>>>>>>> csc - sin
I think the OP means: $\displaystyle \sec x + 2\tan x = \frac{\cot x + 2\cos x}{\csc x - \sin x}$
Start from the RHS. Convert everything into terms of $\displaystyle \sin x$ and $\displaystyle \cos x$. Then multiply both top and bottom by $\displaystyle \sin x$. Split the resulting fraction and simplify.
Hello, speedy229!
I had to guess at what you meant.
And be careful . . . $\displaystyle \sin$ has no meaning
. . . . . just as meaningless as $\displaystyle \sqrt{\;\;}$ or $\displaystyle (\;)^2$
$\displaystyle \sec x+2\tan x \:=\: \frac{\cot x + 2\cos x}{\csc x - \sin x}$
The right side is: .$\displaystyle \frac{\dfrac{\cos x}{\sin x} + 2\cos x}{\dfrac{1}{\sin x} - \sin x} $
Multiply by $\displaystyle \frac{\sin x}{\sin x}\!:\quad \frac{\sin x\left(\dfrac{\cos x}{\sin x} + 2\cos x\right)}{\sin x\left(\dfrac{1}{\sin x} - \sin x\right)} \;=\;\frac{\cos x + 2\sin x\cos x}{1-\sin^2\!x}$ .$\displaystyle = \;\frac{\cos x(1 + 2\sin x)}{\cos^2\!x}$
. . . . . $\displaystyle = \;\frac{1+2\sin x}{\cos x} \;\;=\;\;\frac{1}{\cos x} + \frac{2\sin x}{\cos x} \;\;=\;\;\sec x + 2\tan x$
Edit: I now see that o_O already gave you the game plan . . .
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