Expert: Paul Klarreich - 6/27/2009

Question
The problem I have is (Sec^2x-Tan^2x+Tan)/Secx= Sinx+Cosx

My instructor mentioned sometimes it is easier to change everything in the problem to sine and cosine which I did giving me:

([1/cos^2x]-[sin^2x/cos^2x]+[sinx/cosx])/(1/cosx)  This is where I get stuck, can I move forward from this, if so how do I continue to work the problem?

Answer
Questioner:   Stephen
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  Proving Trigonometric Identities from Pre-Calculus
Question:  The problem I have is (Sec^2x-Tan^2x+Tan)/Secx= Sinx+Cosx

My instructor mentioned sometimes it is easier to change everything in the problem to sine and cosine
.............
>> 'sometimes' is the operative word here.  It won't make it simpler, but it could relieve you of the necessity of learning identities involving tangent, secant, etc.  But that could be useful, too.  Anyway...
............
which I did giving me:

([1/cos^2x]-[sin^2x/cos^2x]+[sinx/cosx])/(1/cosx)  This is where I get stuck, can I move forward from this, if so how do I continue to work the problem?
...............................
Hi, Stephen,

I'm going to abbreviate:

([1/cos^2x]-[sin^2x/cos^2x]+[sinx/cosx])/(1/cosx)  will be written:

1/c^2 - s^2/c^2 + s/c
----------------------
     1/c

This is a 'complex fraction'; you clear denominators -- multiply each term by the LCD = c^2:

c^2(1/c^2) - c^2(s^2/c^2) + c^2(s/c)
------------------------------------- =
      c^2(1/c)

1 - s^2 + sc
---------------
   c

Now  1 - s^2 = c^2:

c^2 + sc
---------
   c

c + s

Hmmm...