How can I do the following questions?

`(costheta)/(1 + sin theta) - costheta/(1 - sin theta) = -2tantheta`

`(costheta(1 - sin theta))/((1 + sin theta)(1 - sin theta)) - (costheta(1 + sin theta))/((1 + sin theta )(1 - sin theta)) =`

`(costheta - costhetasintheta - costheta- costhetasintheta)/((1 + sin theta)(1 - sin theta)) =`

`(-2costhetasintheta)/((1 + sin theta)(1 - sin theta))=`

`(-2costhetasintheta)/(1 - sin^2theta)=`

Use the pythagorean identity `cos^2theta = 1- sin^2theta`.

`(-2costhetasintheta)/(cos^2theta)=`

`(-2sintheta)/costheta =`

Now apply the quotient identity `sintheta/costheta = tantheta`.

`-2tantheta =`

Identity proved!

Hopefully this helps!

Note that `67 1/2˚` is exactly half of `135˚`. This is especially relevant because we know the value of `tan(135˚)`, since `135˚` has a reference angle of `45˚` and is located in the second quadrant, thus `tan(135˚)=-1`.

Looking at the given identity:

`tan(2theta)=(2tan(theta))/(1-tan^2(theta))`

If we let `theta=67 1/2˚`, then we see that

`tan(2xx67 1/2˚)=(2tan(67 1/2˚))/(1-tan^2(67 1/2˚))`

`tan(135˚)=(2tan(67 1/2˚))/(1-tan^2(67 1/2˚))`

We already know the value of `tan(135˚)`:

`-1=(2tan(67 1/2˚))/(1-tan^2(67 1/2˚))`

This will be easier to look at if we let `u=tan(67 1/2˚)`:

`-1=(2u)/(1-u^2)`

Cross-multiply. We want to solve for `u`, since `u=tan(67 1/2˚)`.

`-1(1-u^2)=2u`

`u^2-1=2u`

Solve like you normally would a quadratic equation (set it equal to `0`):

`u^2-2u-1=0`

You could use the quadratic formula here, but I'll complete the square:

`u^2-2u=1`

We want the left side to match `u^2-2u+1=(u-1)^2`. Add `1` to both sides:

`u^2-2u+1=1+1`

`(u-1)^2=2`

Take the square root of both sides:

`u-1=+-sqrt2`

`u=1+-sqrt2`

Since `u=tan(67 1/2˚)`:

`tan(67 1/2˚)=1+-sqrt2`

However, something's up... what should we do about the plus or minus sign? The tangent of a single angle can only equal one thing.

Since `67 1/2˚` is in the first quadrant, we know its tangent must be positive. Thus, we take the only positive solution out of the two:

`tan(67 1/2˚)=1+sqrt2`