Question 168554
): I need help verifying these identities.
 
<pre><font size = 4 color = "indigo"><b>
{{{Sin(x)Cos(x)+Cos^2x = Cos(x)(1+Cot(x))/Csc(x)}}}
 
Work with the right 
side.  Substitute
{{{Cos(x)/Sin(x)}}}
for {{{Cot(x)}}} and
substitute {{{1/Sin(x)}}}
for {{{Csc(x)}}}
 
                      = {{{(Cos(x)(1+Cos(x)/Sin(x)))/(1/Sin(x))}}}
Distribute out the
numerator:
                      = {{{ (Cos(x)+Cos^2x/Sin(x))/(1/Sin(x))}}}
 
Put a 1 under the 
{{{Cos(x)}}} so everything
will be a fraction 
                      = {{{ (Cos(x)/1+Cos^2x/Sin(x))/(1/Sin(x))}}}


Invert the fraction
in the denominator 
and multiply:
 
                      = {{{ (Cos(x)/1+Cos^2x/Sin(x) )*( Sin(x)/1 )}}}
 
Swap the fraction factors:

                      = {{{( Sin(x)/1 )*(Cos(x)/1+Cos^2x/Sin(x) )}}}                   
 
 
Distribute:

                      = {{{( Sin(x)/1 )*(Cos(x)/1)+(Sin(x)/1)*(Cos^2x/Sin(x)) )}}} 

                      = {{{(Sin(x)Cos(x)/1)+(cross(Sin(x))/1)*(Cos^2x/cross(Sin(x))) )}}}

                      = {{{Sin(x)Cos(x)+Cos^2x}}}

AND 
{{{(Sec(x)-Tan(x))^2 + (Sec(x)+Tan(x))^2 = 2+4Tan^2x}}}

Work with the left side. 
Square out the two 
binomials:

{{{(Sec(x)-Tan(x))^2 + (Sec(x)+Tan(x))^2}}} =

{{{matrix(1,3,Sec^2x-2Sec(x)Tan(x)+Tan^2x,"+", Sec^2x+2Sec(x)Tan(x)+Tan^2x )}}}

{{{matrix(1,3,Sec^2x-cross(2Sec(x)Tan(x))+Tan^2x,"+", Sec^2x+cross(2Sec(x)Tan(x))+Tan^2x )}}}
 
{{{2Sec^2x + 2Tan^2x}}}

Use the identity {{{1+Tan^2alpha=Sec^2alpha}}} and
replace {{{Sec^2x}}} by {{{(1+Tan^2x)}}}

{{{ 2(1+Tan^2x) + 2Tan^2x}}}

Distribute:

{{{ 2+2Tan^2x+2Tan^2x }}}

Combine like terms:

{{{2+4Tan^2x}}}

Edwin</pre>