# How can I do the following questions?

## How do you find the value of tan(67.5˚) How do you prove that $\cos \frac{\theta}{1 + \sin \theta} - \cos \frac{\theta}{1 + \sin \theta} = - 2 \tan \theta$?

Jun 13, 2016

Start by putting the left side on a common denominator.

#### Explanation:

$\frac{\cos \theta}{1 + \sin \theta} - \cos \frac{\theta}{1 - \sin \theta} = - 2 \tan \theta$

$\frac{\cos \theta \left(1 - \sin \theta\right)}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} - \frac{\cos \theta \left(1 + \sin \theta\right)}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} =$

$\frac{\cos \theta - \cos \theta \sin \theta - \cos \theta - \cos \theta \sin \theta}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} =$

$\frac{- 2 \cos \theta \sin \theta}{\left(1 + \sin \theta\right) \left(1 - \sin \theta\right)} =$

$\frac{- 2 \cos \theta \sin \theta}{1 - {\sin}^{2} \theta} =$

Use the pythagorean identity ${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$.

$\frac{- 2 \cos \theta \sin \theta}{{\cos}^{2} \theta} =$

$\frac{- 2 \sin \theta}{\cos} \theta =$

Now apply the quotient identity $\sin \frac{\theta}{\cos} \theta = \tan \theta$.

$- 2 \tan \theta =$

Identity proved!

Hopefully this helps!

Jun 14, 2016

Solving number 30...

#### Explanation:

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:

$\tan \left(2 \theta\right) = \frac{2 \tan \left(\theta\right)}{1 - {\tan}^{2} \left(\theta\right)}$

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˚):

$- 1 = \frac{2 u}{1 - {u}^{2}}$

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

$- 1 \left(1 - {u}^{2}\right) = 2 u$

${u}^{2} - 1 = 2 u$

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

${u}^{2} - 2 u - 1 = 0$

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

${u}^{2} - 2 u = 1$

We want the left side to match ${u}^{2} - 2 u + 1 = {\left(u - 1\right)}^{2}$. Add $1$ to both sides:

${u}^{2} - 2 u + 1 = 1 + 1$

${\left(u - 1\right)}^{2} = 2$

Take the square root of both sides:

$u - 1 = \pm \sqrt{2}$

$u = 1 \pm \sqrt{2}$

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