How can I do the following questions?

  1. How do you find the value of #tan(67.5˚)#

  2. How do you prove that #costheta/(1 + sin theta) - costheta/(1+sintheta) = -2tantheta#?

2 Answers
Noah G · mason m
Jun 13, 2016

Start by putting the left side on a common denominator.

Explanation:

#(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!

mason m
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(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#

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