Verifying an identity

[coooool222]

[math]1/tanx - sec x + 1/tan x + sec x = -2tanx[/math]
I'm having a little difficulty with this

The reciprocal of 1/tan x - sec x is
cos x / sin x - cos

The reciprocal of 1/tax + sec x is
cos x/ sin x + cos

When i add these together
cos x / sin x - cos + cos x / sin x + cos x
i get 2 cos / sin x
which is then 2 cot? so how is it -2tanx

 

[skeeter]

you really need to use parentheses around the terms in the respective denominators to make your expression clear …

[math]\dfrac{1}{\tan{x}-\sec{x}} = \dfrac{\cos{x}}{\sin{x}-1}[/math]
[math]\dfrac{1}{\tan{x}+\sec{x}} = \dfrac{\cos{x}}{\sin{x}+1}[/math]
common denominator is [imath](\sin{x}-1)(\sin{x}+1)=\sin^2{x}-1 = -\cos^2{x}[/imath]

try adding the two expressions again …

 

[coooool222]

wait how come its cos x / sin x -1 not Cos/sin x - cos

 

[skeeter]

multiply the two fractions …

[math]\dfrac{\cos{x}}{\cos{x}} \cdot \dfrac{1}{\tan{x}-\sec{x}}[/math]
… what do you get?

 

[coooool222]

why are you multiplying with cos x / cos x isnt that 1 also why cant i do sin x / sin x

 

[skeeter]

why are you multiplying with cos x / cos x isnt that 1

yes … when you multiply any expression by 1, don’t you get the same expression in a different form?

 

[skeeter]

… why cant i do sin x / sin x

you can, but it would make adding the two original fractions more difficult

 

[skeeter]

why are you multiplying with cos x / cos x

it clears the complex fraction …

[math]\dfrac{\cos{x}}{\cancel{\cos{x}}} \cdot \dfrac{1}{\dfrac{\sin{x}}{\cancel{\cos{x}}} - \dfrac{1}{\cancel{\cos{x}}}} = \dfrac{\cos{x}}{\sin{x}-1}[/math]