Question

Trig help (Problems already solved)?

The calculation for the problems are already done but I have to list a reason or what is being done in each step. "Each = and newline made" means a place I have to write what is being done in the calculation.

1. (secx + sinx)cotx = cscx + cosx
=(secx + sinx)cotx = cscx + cosx
=(1 / sinx) + cosx
=cscx + cosx

2. cosx + tanx sinx = secx
=cosx + tanx sinx = cosx + (sinx / cosx)sinx
=cosx + (sin^2x / cosx) = (1 / cosx)(cos^2x + sin^2x)
=1 / cosx
=secx

3. cscx - cosx cotx = sinx
=cscx - cosx cotx = (1 / sinx) - cosx(cosx / sinx)
=(1 / sinx) - (cos^2x / sinx)
=(1 - cos^2x) / sinx
=sin^2x / sinx = sinx

4. (cosx / (1 + cosx)) + (cosx / (1 - cosx)) = 2cotx cscx
=(cosx / (1 + cosx)) + (cosx / (1 - cosx)) = ((cosx (1 - cosx) + cosx (1 + cosx))) / (1 + cosx)(1 - cosx)
=(cosx - cos^2x + cosx + cos^2x) / (1 - cos^2x)
=2cosx / sin^2x
=2(cosx / sinx)(1 / sinx) = 2cotx cscx

Thank you to whoever decides to help me with explaining what is happening on each line.

Answer 1

The first identity uses the definition of the reciprocal functions
`\sec x,\csc x,\cot x` and the distributive property of multiplication.


`(\sec x+\sin x)\cot x=\left(\frac1{\cos x}+\sin x\right)(\cos x)/(\sin x)`

`=(\cos x)/(\cos x\sin x)+(\cos x\sin x)/(\sin x)`

`=\frac1{\sin x}+\cos x`

`=\csc x+\cos x`

The second uses the definition of
`\tan x` and the distributive property. Then a factor of
`\frac1{\cos x}` is pulled out, which allows you to use the identity
`\sin^2x+\cos^2x=1`.


`\cos x+\tan x\sin x=\cos x+(\sin x)/(\cos x)\sin x`

`=\cos x+(\sin^2x)/(\cos x)`

`=(\cos^2x)/(\cos x)+(\sin^2x)/(\cos x)`

`=\frac1{\cos x}\left(\cos^2x+\sin^2x\right)`

`=\frac1{\cos x}*1`

`=\frac1{\cos x}`

`=1`

The third uses the same ideas as the second: rewrite the reciprocal functions, then invoke the Pythagorean identity
`\sin^2x+\cos^2x=1`, which is equivalent to
`\sin^2x=1-\cos^2x`.


`\csc x-\cos x\cot x=\frac1{\sin x}-\cos x(\cos x)/(\sin x)`

`=\frac1{\sin x}-(\cos^2x)/(\sin x)`

`=\frac1{\sin x}\left(1-\cos^2x\right)`

`=\frac1{\sin x}\sin^2x`

`=(\sin^2x)/(\sin x)`

`=\sin x`

In the last one, you combine the fractions by enforcing common denominators. This lets you add the numerators together, and the denominator can be simplified. Once you do that, you rewrite the factors of cos and sin in the numerator and denominator to make up the cot and csc functions, and you're done.


`(\cos x)/(1+\cos x)+(\cos x)/(1-\cos x)=(\cos x(1-\cos x))/((1+\cos x)(1-\cos x))+(\cos x(1+\cos x))/((1-\cos x)(1+\cos x))`

`=(\cos x(1-\cos x)+\cos x(1+\cos x))/((1-\cos x)(1+\cos x))`

`=(\cos x(1-\cos x+1+\cos x))/(1-\cos^2x)`

`=(2\cos x)/(\sin^2x)`

`=2(\cos x)/(\sin x)\frac1{\sin x}`

`=2\cot x\csc x`