Expert: Scotto - 4/18/2009

Question
Hello! I'm having trouble with a couple of questions. Can
you please help me?

1. sinx/1-cosx + 1-cosx/sinx = 2cscx
2. cosx/secx + sinx/cscx = sec^2x - tan^2x
3. secx/secx-tanx = sec^2x + secxtanx
4. 1/1-sinx + 1/1+sinx = 2sec^2x
5. tanx/1+cosx + sinx/1-cosx = cotx + secxcscx

Thank you!

Answer
1. Get a common denominator.  That gives
(sin²x + (1-cosx)²) / (sinx(1-cosx)).

Numerator

Hold on to that thought while we take a moment to look at the second term in the numeratior, the (1-cosx)².
It can be seen that (1-cosx)² =  1 - 2cosx + cos²x.

Now back to the other term in the top, sin²x.
Add this to what we just got.
Since we have sin²x + 1 - 2cosx + cos²x in the numerator,
it is known that sin²x + cos²x = 1, so this is really
1 + 1 + - 2cosx = 2 - 2cosx = 2(1-cosx).

Denominator

Now if the denominator is looked at, it can be seen to be
sinx(1-cosx).

Both
The 1-cosx in the numerator and the denominator cancel,
leaving us with 2 in the numerator and sinx in the denominator.
It is also known the 1/sinx = cscx, so we have 2cscx.
That looks like what we're looking for.


2. In this problem, again I'll start with the left side.
That's cosx/secx + sinx/cscx.
It is known that 1/secx = cosx and 1/cscx = sinx,
so this is cos²x + sin²x.  That is known to be 1.

It is also known that 1 + tan²x = sec²x,
so this can be rewritten as 1 = sec²x - tan²x.

Substitute that for 1, and we have the answer.


3. Looking at secx/(secx - tanx), multiply the top and bottom by
secx + tanx.  That gives sex²x + secxtanx on top.
It gives sec²x - tan²x on the bottom.
Since we know that 1 + tan²x = sec²x,
it can be shown that sec²x - tan²x = 1, so the bottom disappears.

All we have left is the top, which is sec²x + secxtanx.
That's it.

4. Getting a common denominator out of the 1-sinx and 1+sinx in the bottom of both the fractions gives us (1+sin + 1-sinx)/(1-sin²x).
The numerator becomes 2 and the denominator is cos²x.
It is known that 1/cos²x = sec²x, so the result is just 2sec²x.

5. We are given tanx/(1+cosx) + sinx/(1-cosx).
Getting a common denominator gives us
(tanx(1-cosx) + sinx(1+cosx))/(1-cos²x).

Denominater
Now it is known that the 1-cos²x in the denominator is really sin²x.

Numerator
In the numerator we have tanx  - tanxcosx + sinx + sinxcosx.

Splitting Fraction
Lets split the fraction into the four pieces.

They are 1) tanx/sin²x, 2) –tanxcosx/sin²x, 3) sinx/sin²x, and 4) sinxcosx/sin²x.

I will look at each of them before putting them back together.

The first one is (sinx/cosx)/sin²x.
The sinx/sinx cancels, leaving 1/(cosxsinx).
That’s secxcscx.

The second one is –(sinx/cosx)cosx(1/sin²x).
The cosx/cosx cancels.
The sinx/sin²x is 1/sinx = cscx.
That leaves –cscx.

The thrid one if sinx/sin²x, which is 1/sinx, which is known to be cscx.

The fourth one is sinxcosx/sin².  We can cancel the sinx/sinx,
Leaving cosx/sinx, which is ctnx.

We can see that (2) and (3) cancel, since (2) is –(3), so all we have is (1) and (4).
That is, secxcscx + ctnx.  Reverse it and the right side is obtained.