10-2

This week I'm going to explain how to do problems from 10-2 since I haven't done 10-2 yet. 10-2 uses two formulas that involves tan. It isn't hard since both of the problems look exactly the same except for the signs changing.

Formulas:

• Tan(a+b)=Tana+Tanb/1-TanaTanb
• Tan(a-b)=Tana-Tanb/1+TanaTanb

******A good way to remember the formulas is that the top sign is the same as the sign in the parentheses and the bottom sign is the opposite of the sign in the parentheses.******

Ex. 1: Tana=5/4 and Tanb=3/7. Solve for Tan(a-b).

• First, you have to plug the Tana and Tanb that it gave you into the formula.
• Tan(a-b)=Tana-Tanb/1+TanaTanb=(5/4)-(3/7)/1+(5/4)(3/7)
• Ok, now you have to solve the equation.
  
• (5/4)-(3/7)/1+(5/4)(3/7)=(35/28)-(12/28)/1+(15/28)
• Ok, so on the top of the equation I multiplied 5/4 by 7/7 and 3/7 by 4/4 because you can't subtract fractions unless they have a common denominator. Then on the bottom, all I did was multiply straight across.
• Continuing with the equation: (35/28)-(12/28)/1+(15/28)=(23/28)/(43/28)
• Now, this time on the bottom I replaced 1 with 28/28 and just added across because you need common denominators to add fractions.
• Continuing with the equation: (23/28)/(43/28)=644/1204=23/43. Final answer is 23/43. For this step, I just multiplied 23/28 by the reciprocal of 43/28 which is 28/43.

If you had to use the other formula for this problem, you solve it the same way. The only that is different are the signs.


-Amber :)