Question 230137
Your expression is {{{sin(x) * tan(x)}}}


Since {{{tan(x) = sin(x) / cos(x)}}}, your expression becomes:


{{{(sin(x) * sin(x))/cos(x))}}}


This becomes:


{{{sin(x)^2 / cos(x)}}}


Since {{{sin(x)^2 + cos(x)^2 = 1}}}, this means that:


{{{sin(x)^2 = 1 - cos(x)^2}}}


Replace {{{sin(x)^2}}} in your expression with this to get:


{{{(1 - cos(x)^2) / cos(x)}}}


Since {{{cos(x) = 1/sec(x)}}}, you can replace cos(x) in your equation to get:


{{{(1 - cos(x)^2) / (1/sec(x))}}}


This is the same as:


{{{(1 - cos(x)^2) * (sec(x))}}} because {{{a/(1/b) = a*b}}}.


Your expression has become:


{{{(1 - cos(x)^2) * sec(x)}}}


Remove parentheses to get:


{{{sec(x) - (cos(x)^2 * sec(x))}}}


Since {{{sec(x) = 1/cos(x)}}}, you can replace in your equation to get:


{{{sec(x) - (cos(x)^2/cos(x))}}}


Simplify to get:


{{{sec(x) - cos(x)}}}


Since this is what you wanted to prove, you're done.


Note I am showing sin^2(x) as sin(x)^2 because it doesn't come out good on the formula rendering routine with sin^2(x).


Example:


sin^2(x) shows up as {{{sin^2(x)}}}


sin(x)^2 shows up as {{{sin(x)^2}}}


The second version is technically incorrect but it shows up clearer so I used it.


Note that sin(x) * sin(x) really is (sin(x))^2 but I shortened it to sin(x)^2 to eliminate all those extra parentheses that muddied up the presentation.


Just remember that sin(x)^2 is the same as sin^2(x) and we'll be ok.