Question 985872
<pre>
1)If sinA=sinB and cosA=cosB then A=B or else they differ by {{{2pi}}} or {{{360°}}}
 

Since the period of sine and cosine is {{{2pi}}} or 360°, the general solution is 
{{{A=B+2pi*n}}} or A = B + 360°n, where n is any integer   

2)If 
{{{(1+tan(x)^"")(1+tan(y)^"")}}}{{{""=""}}}{{{2}}} then {{{x+y}}}{{{""=""}}}_____.

FOIL out the left side:
{{{1 + tan(x) + tan(y) + tan(x)tan(y)}}}{{{""=""}}}{{{2}}}
If we remember our identities and also notice that we are 
asked to find x+y, and notice that there are tangents in
the problem and then we we think of the identity we 
have memorized for tan(x+y)
{{{tan(x+y)}}}{{{""=""}}}{{{(tan(x)+tan(y))/(1-tan(x)tan(y))}}}
We recognize some of the terms in our equation are like some
of the terms in that identity so let's get tan(x)+tan(y) alone
on the left side since that's the numerator of the right side
of that identity:
{{{1 + tan(x) + tan(y) + tan(x)tan(y)}}}{{{""=""}}}{{{2}}}

{{{tan(x) + tan(y)}}}{{{""=""}}}{{{1-tan(x)tan(y)}}}

Well what do you know! the right side is just the denominator 
of the right side of that identity: 
So we divide both sides by the right side:    
{{{(tan(x) + tan(y))/(1-tan(x)tan(y))}}}{{{""=""}}}{{{(1 - tan(x)tan(y))/(1-tan(x)tan(y))}}} 
  
The left side is the left side of that identity and the right side is 1. 
So we have

{{{tan(x+y)}}}{{{""=""}}}{{{1}}}
Since the period of the tangent is {{{pi}}} or 180°
x+y = {{{pi/4+pi*n}}} or 45°+180°n, where n is any integer, positive,
negative, or zero. 

Limit: 2 problems

Edwin</pre>