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? {{{ (cot^2(a))/(csc^2(a)-1)=(1+sin(a))/(sin(a)) }}}
<pre><font size = 4 color = "indigo"><b>
This is not an identity. If it were it would be
true for every angle a.

Suppose we had this right triangle:

{{{drawing(200,200,-1,5,-1,5, triangle(0,0,3,0,3,4), 
locate(1.5,0,3), locate(3.1,2,4), locate(.4,.5,a) 

)}}} 

Calculating its hypotenuse

{{{c^2=a^2+b^2}}}
{{{c^2=3^2+4^2}}}
{{{c^2=9+16}}}
{{{c^2=25}}}
{{{c=sqrt(25)}}}
{{{c=5}}}

{{{drawing(200,200,-1,5,-1,5, triangle(0,0,3,0,3,4), 
locate(1.5,0,3), locate(3.1,2,4), locate(.4,.5,a), 
locate(1.1,2.4,5)
)}}}


Substituting the trig ratios from angle a from
this triangle,


 {{{ matrix(6,3,

 (Cot^2a)/(Csc^2a-1), "=",(1+Sin(a))/(Sin(a))   ,

(3/4)^2/((5/4)^2-1), "=", (1+4/5)/(4/5),

(.75)^2/((1.25)^2-1), "=", (1+.8)/(.8),

.5625/(1.5625-1), "=", 1.8/.8,

.5625/.5625, "=", 2.25,
 
1, "=", 2.25 ) }}}

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

Edwin</pre>