Expert: Chen Min - 1/27/2011

Question
Hi Chen,

I'm wondering if you could guide me through another Trig proving question that's really giving me trouble.  I don't think it can be proved...what do you think?  Here's the question:

tan(x)-1 = sin^2(x)-cos^2(x) *over*
          sin(x)cos(x)+cos^2(x)

Please note that by *over* I mean that those values are to be represented as a fraction.  Any help you can give me with this one would be so awesome!!

Thanks so much, Oliver

Answer
RHS=(sinx+cosx)(sinx-cosx)/cosx(sinx+cosx)
  =(sinx-cosx)/cosx
  =tanx - 1

This one is actually easy

For more complicated ones, just use the following equations:

Imagine a hexagon, with top two apex reading sin and cos, middle two apex reading tan and cot, the bottom two reading sec and csc. Thus six trig functions are at six points now. For every diagonal, the two things at each end are reciprocal of each other, e.g. sinx=1/cscx, tanx=1/cotx.

Now, divide the hex into six identical triangles, each trigo function has two triangles touching it. Now, assign the middle point that touches all six triangle the value 1

For every upside-down triangle, the sum of the squares of the top two equals the bottom's sqaure: sin^2x+cos^x=1^2, tan^2x + 1 = sec^2 x.

Know and use well all these conversions and you'll be good for elementary trigos