SOLUTION: Please help me with this equation. I have to use Trig. Identities to solve it and i can't seem to get both sides to equal the same thing. Could you show me how to get there? {{{ (c

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me with this equation. I have to use Trig. Identities to solve it and i can't seem to get both sides to equal the same thing. Could you show me how to get there? {{{ (c      Log On


   

Question 176512: Please help me with this equation. I have to use Trig. Identities to solve it and i can't seem to get both sides to equal the same thing. Could you show me how to get there? +%28cot%5E2%28a%29%29%2F%28csc%5E2%28a%29-1%29=%281%2Bsin%28a%29%29%2F%28sin%28a%29%29+
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me with this equation. I have to use Trig. Identities to solve it and i can't seem to get both sides to equal the same thing. Could you show me how to get there? +%28cot%5E2%28a%29%29%2F%28csc%5E2%28a%29-1%29=%281%2Bsin%28a%29%29%2F%28sin%28a%29%29+

This is not an identity. If it were it would be
true for every angle a.

Suppose we had this right triangle:

 

Calculating its hypotenuse

c%5E2=a%5E2%2Bb%5E2
c%5E2=3%5E2%2B4%5E2
c%5E2=9%2B16
c%5E2=25
c=sqrt%2825%29
c=5




Substituting the trig ratios from angle a from
this triangle,


 

So it is not an identity.  Somebody made a typo,
and it may very well have been the book publisher.

Edwin