Prove ((sec(x)-tan(x))^2 + 1)/csc(x)(sec(x)-tan(x)) = 2tan(x

[soulofeternity]

In order to prove the statement true, you must simplify one side down until you have the same thing on each side (I hope that made sense . . .). If that didn't make sense, it just means simplify the left side to 2tan(x).

((sec(x)-tan(x))^2 + 1)/ csc(x)(sec(x)-tan(x)) = 2tan(x)

 

[skeeter]

[(secx - tanx)² + 1]/[cscx(secx - tanx)] =

(sec²x - 2secxtanx + tan²x + 1)/[cscx(secx - tanx)] =

(2sec²x - 2secxtanx)/[cscx(secx - tanx)] =

2secx(secx - tanx)/[cscx(secx - tanx)] =

2secx/cscx =

2sinx/cosx =

2tanx

 

[soroban]

Re: Proving Identities

Hello, soulofeternity!

$\displaystyle \:\frac{[\sec(x)\,-\,\tan(x)]^2\,+\,1}{\csc(x)[\sec(x)\,-\,\tan(x)]}\; =\; 2\tan(x)$

The numerator is: $\displaystyle \:[\sec(x)\,-\,\tan(x)]^2\,+\,1$

. . $\displaystyle =\:\sec^2(x)\,-\,2\sec(x)\tan(x)\,+\,\underbrace{\tan^2(x)\,+\,1}$
. . $\displaystyle = \:\sec^2(x)\;-\;2\sec(x)\tan(x)\;+\;\sec^2(x)$

. . $\displaystyle = \:2\sec^2(x)\,-\,2\sec(x)\tan(x)$

. . $\displaystyle = \;2\sec(x)\cdot[\sec(x)\,-\,\tan(x)]$


The function becomes: \(\displaystyle \L\:\frac{2\sec(x)\cdot[\sec(x)\,-\,\tan(x)]}{\csc(x)\cdot[\sec(x)\,-\,\tan(x)] } \:=\:\frac{2\sec(x)}{\csc(x)} \:=\:\frac{2\sin(x)}{\cos(x)}\)$\displaystyle \:=\:2\tan(x)$

Too fast for me, skeeter!