by 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)`
Explore and know. That is asked.