Expert: Scott A Wilson - 11/2/2011

Question
i have two questions im hoping you can walk me through with.... thy are both attached and i have been working at these for several hour between last night and this moning so if you could please show me how to ddo this or do it for me it would be VERRRYYYY appreciated!!!.

Answer
On my PC, the problems come across very tiny, so I'm not sure I read them correctly
for the text is just a little blurry.

The 1st one looks like simplify cos²x * sec²x - cos²x.
Factoring out a cos²x gives cos²x(sec²x - 1).

Since we know 1 + tan²x = sec²x, sec²x - 1 = tan²x.
This makes it be cos²x * tan²x.

Since we know that tanx = sinx/cosx, this makes tan²x = sin²x/cos²x.
Putting this in gives cos²x * sin²x / cos²x.
Cancelling like terms leaves us with sin²x, which looks like (D).


For the 2nd problem, it looks like there is a 1-sinx in the numerator and the denominator,
and these cancel.  It looks like what is left is cosx/ctnx, and this is just sinx,
since cosx/ctnx = cosx*tanx = cosx*sinx/cosx = sinx, which looks like (A).