trig proof road block

**speedy229** · November 15th 2008, 06:27 PM

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

**11rdc11** · November 15th 2008, 06:45 PM

Originally Posted by speedy229

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'm kind of confused what you are asking. Are you asking to prove?

$\sec{x} +2\tan{x} = \cot{x} + 2\cos{x}$

If so what relevance does

$\csc{x} - \sin{x}$

have to the problem?

**o_O** · November 15th 2008, 07:22 PM

I think the OP means: $\sec x + 2\tan x = \frac{\cot x + 2\cos x}{\csc x - \sin x}$

Start from the RHS. Convert everything into terms of $\sin x$ and $\cos x$ . Then multiply both top and bottom by $\sin x$ . Split the resulting fraction and simplify.

**Soroban** · November 15th 2008, 07:32 PM

Hello, speedy229!

I had to guess at what you meant.

And be careful . . . $\sin$ has no meaning
. . . . . just as meaningless as $\sqrt{\;\;}$ or $(\;)^2$

$\sec x+2\tan x \:=\: \frac{\cot x + 2\cos x}{\csc x - \sin x}$

The right side is: . $\frac{\dfrac{\cos x}{\sin x} + 2\cos x}{\dfrac{1}{\sin x} - \sin x}$

Multiply by $\frac{\sin x}{\sin x}\!:\quad \frac{\sin x\left(\dfrac{\cos x}{\sin x} + 2\cos x\right)}{\sin x\left(\dfrac{1}{\sin x} - \sin x\right)} \;=\;\frac{\cos x + 2\sin x\cos x}{1-\sin^2\!x}$ . $= \;\frac{\cos x(1 + 2\sin x)}{\cos^2\!x}$

. . . . . $= \;\frac{1+2\sin x}{\cos x} \;\;=\;\;\frac{1}{\cos x} + \frac{2\sin x}{\cos x} \;\;=\;\;\sec x + 2\tan x$

Edit: I now see that o_O already gave you the game plan . . .
.

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