cos x + sin x tan x and `tan x/ (sec x - cos x)`

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cos x + sin x tan x and `tan x/ (sec x - cos x)`

Postby Galaxy » Mon 11 April, 2011 19:00

Write the trigonometric expression in terms of sine and cosine, and then simplify.
1. `cos x + sin x tan x ` and
2. `tan x/ (sec x - cos x)`
Last edited by Galaxy on Mon 11 April, 2011 20:06, edited 1 time in total.
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Re: cos x + sin x tan x and `tan x/ (sec x - cos x)`

Postby barnamah » Mon 11 April, 2011 19:49

for question 1
for the ` cos x + sin x tan x` first write `tanx=sinx/cosx` so we have:

` cos x + sin x sinx/cosx=cosx+(sin^2x)/cosx` taking cosx common we have:

`cosx+(sin^2x)/cosx=(cos^2x+sin^2x)/cosx` and w know `sin^2x+cos^2x=1` so we have:
`1/cosx=secx`

for question 2:
`tan x/ (sec x - cos x)`
lets take care of the denominator `sec x - cos x`
we know `secx=1/cosx` so we can write it as `1/cosx-cosx` using cosx as common denominator we have;
`1/cosx-cosx=(1-cos^2x)/cosx`

Putting it into our equation

`tan x/ (sec x - cos x)=(tanx)/((1-cos^2x)/cosx)`
now tanx `-:` by the `(1-cos^2x)/cosx`
`tanx-:(1-cos^2x)/cosx`

so we can flip the `(1-cos^2x)/cosx` and change the sin from `-:` to multiplication and we have:

`tanx-:(1-cos^2x)/cosx=tanx*(cosx)/(1-cos^2x)`

As know that `tanx=sinx/cosx` so we use this in our equation and we have:
`tanx*(cosx)/(1-cos^2x)=sinx/cosx*(cosx)/(1-cos^2x)`
The cosx at top and bottom cancels out and have:
`sinx/(1-cos^2x)`

we know tha `sin^2+cos^2=1` by moving cosx to the right side we have `sinx^2x=1-cos^2x` so in our equation we can use `sin^2x` for `1-cos^2x`
`sinx/(1-cos^2x)=sinx/(sin^2x)=1/sinx`

we know`1/sinx=csc(x)`

`:.` the final answer is `csc(x)`
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