Verify Trig. Identity: cos(-x) / 1+tan(-x) - sin(-x) / 1+cot

[rebmik]

I am having trouble with another one.

cos(-x) / 1+tan(-x) - sin(-x) / 1+cot(-x) = sinx + cosx

I started by trying to use the even/odd identities to change (-x) to (x), then I changed everything to sin and cos, then tried a LCD. Now I have a big hugh mess.

Thank you for any help you can give me.

 

[skeeter]

please ... learn to use proper grouping symbols to make your expressions mathematically clear.


cos(-x)/[1+tan(-x)] - sin(-x)/[1+cot(-x)] =

use odd/even identities ...

cos(x)/[1-tan(x)] + sin(x)/[1-cot(x)] =

multiply 1st fraction by cosx/cosx, 2nd fraction by sinx/sinx ...

cos²x/(cosx-sinx)+sin²x/(sinx-cosx) =

cos²x/(cosx-sinx)-sin²x/(cosx-sinx) =

(cos²x-sin²x)/(cosx-sinx) =

(cosx+sinx)(cosx-sinx)/(cosx-sinx) =

cosx+sinx

 

[soroban]

Re: Verify Trig. Identity: cos(-x) / 1+tan(-x) - sin(-x) / 1

Hello, rebmik!

\(\displaystyle \L\frac{\cos(-x)}{1\,+\,\tan(-x)}\,-\,\frac{\sin(-x)}{1\,+\,\cot(-x)} \:=\:\sin x\,+\,\cos x\)

I started by trying to use the even/odd identities to change (-x) to (x),
then I changed everything to sin and cos, then tried a LCD.
This should have worked, looking like Skeeter's solution.

We have: \(\displaystyle \L\:\frac{\cos(-x)}{1\,+\,\tan(-x)}\,-\,\frac{\sin(-x)}{1\,+\,\cot(-x)} \;= \;\frac{\cos x}{1\,-\,\tan x}\,+\,\frac{\sin x}{1\,-\,\cot x} \;=\; \frac{\cos x}{1\,-\,\frac{\sin x}{\cos x}}\,+\,\frac{\sin x}{1\,-\,\frac{\cos x}{\sin x}}\)


Multiply the first fraction by \(\displaystyle \frac{\cos x}{\cos x}\), the second fraction by \(\displaystyle \frac{\sin x}{\sin x}\)

. . \(\displaystyle \L\frac{\cos x}{cos x}\,\cdot\,\left(\frac{\cos x}{1\,-\,\frac{\sin x}{\cos x}}\right) \,+\,\frac{\sin x}{\sin x}\,\cdot\,\left(\frac{\sin x}{1\,-\,\frac{\cos x}{\sin x}}\right) \;= \; \frac{\cos^2x}{\cos x\,-\,\sin x}\,+\,\frac{\sin^2x}{\sin x\,-\,\cos x}\)

. . \(\displaystyle \L= \;\frac{\cos^2x}{\cos x\,-\,\sin x} \,- \,\frac{\sin^2x}{\cos x\,-\,\sin x} \;=\;\frac{\cos^2x\,-\,\sin^2x}{\cos x\,-\,\sin x}\)


Factor: \(\displaystyle \L\;\frac{(\sout{\cos x\,-\,\sin x})(\cos x\,+\,\sin x)}{\sout{\cos x\,-\,\sin x}}\)\(\displaystyle \;=\;\cos x\,+\,\sin x\)

 

[rebmik]

Hi Skeeter & Soroban.

Sorry about the grouping symbols.

I was messing up on the multiplication, for some reason I was thinking you had to multiply both by the same thing so I was trying to multi both by sin and that was goofing me up.

Thanks for clearing that up.

Have a nice weekend.