Question 552116
<pre>
1 + tan(A)·tan{{{(A/2)}}} = tan(A)cot{{{(A/2)}}} - 1 = sec(A)

We will make use of the half angle identities:

tan{{{(A/2)}}} = {{{(1-cos(A))/sin(A)}}}, and its reciprocal cot{{{(A/2)}}} = {{{sin(A)/(1-cos(A))}}}, also tan(A) = {{{sin(A)/cos(A)}}} and sec(A) = {{{1/cos(A)}}}

First we will prove:

1 + tan(A)·tan{{{(A/2)}}} = sec(A)

1 + {{{(sin(A)/cos(A))}}}{{{((1-cos(A))/sin(A))}}} 

1 + {{{   (sin(A)-sin(A)cos(A))  /(cos(A)sin(A))   }}}

Get an LCD

{{{(cos(A)sin(A))/(cos(A)sin(A))}}} + {{{   (sin(A)-sin(A)cos(A))  /(cos(A)sin(A))   }}}

Combine fractions over the LCD

{{{(cos(A)sin(A) +    sin(A)-sin(A)cos(A))  /(cos(A)sin(A))   }}}

Cancel first and third terms on top:

{{{(cross(cos(A)sin(A)) +    sin(A)-cross(sin(A)cos(A)))  /(cos(A)sin(A))   }}}

{{{sin(A)  /(cos(A)sin(A))   }}}

Cancel the sines:

{{{cross(sin(A))  /(cos(A)cross(sin(A)))   }}}

{{{1/cos(A)}}}

sec(A)

---------------------------------

Next we will prove:

tan(A)·cot{{{(A/2)}}} - 1 = sec(A)

{{{(sin(A)/cos(A))}}}{{{(sin(A)/(1-cos(A)))}}} - 1 

{{{   (sin^2(A))  /(cos(A)(1-cos(A)))   }}} - 1

Get an LCD

{{{   (sin^2(A))  /(cos(A)(1-cos(A)))   }}} - {{{   (cos(A)(1-cos(A)))  /(cos(A)(1-cos(A)))   }}}

Combine fractions over the LCD

{{{   (sin^2(A)   -    cos(A)(1-cos(A)))  /(cos(A)(1-cos(A)))   }}}

Distribute:

{{{   (sin^2(A)   -    cos(A)+cos^2(A))  /(cos(A)(1-cos(A)))   }}}

Rearrange:

{{{   (sin^2(A)+cos^2(A)-cos(A))  /(cos(A)(1-cos(A)))   }}}

Use identity sin²(A)+cos²(A)=1

{{{   (1-cos(A))  /(cos(A)(1-cos(A)))   }}}

Cancel 1-cos(A)'s

{{{   (cross(1-cos(A)))  /(cos(A)(cross(1-cos(A))))   }}}

{{{1/cos(A)}}}

sec(A)

Edwin</pre>