First, it would be nice if we could put all of our angle measures in terms of a single variable. Since we're dealing with the interior angles of a triangle, and since the interior angles of a triangle add to 180°, we can say that
A + B + C = 180, or equivalently,
B + C = 180 - A
Knowing this, we can replace the B + C on the left side of the equation with 180 - A so that our equation reads:
[tex] \sec^2\big({\frac{180-A}{2}\big)-1=\cot^2\big({\frac{A}{2} }\big) [/tex]
Doing a little tidying, we can rewrite the term [tex] \frac{180-A}{2} [/tex] as
[tex] \dfrac{180-A}{2} =\dfrac{180}{2}-\dfrac{A}{2} =90-\frac{A}{2} [/tex]
By definition, [tex] \sec^2x=\frac{1}{\cos^2x} [/tex], so I'd like to rewrite the left side of the equation with these last two things in mind:
[tex] \sec^2\big({\frac{180-A}{2}\big)-1=\dfrac{1}{\cos^2(90-\frac{A}{2})} -1 [/tex]
[tex] \cos{(90-x)}=\sin{x} [/tex], so we can rewrite the left side again as
[tex] \dfrac{1}{\sin^2\big(\frac{A}{2}\big)} -1 [/tex]
Next, we note that - by definition, [tex] \cot{x}=\frac{\cos{x}}{\sin{x}} [/tex]. This allows us the rewrite the right side of the equation as
[tex] \cot^2\big(\frac{A}{2}\big)=\dfrac{\cos^2\big(\frac{A}{2}\big)}{\sin^2\big({\frac{A}{2}}\big)} [/tex]
And our full equation as
[tex] \dfrac{1}{\sin^2\big(\frac{A}{2}\big)} -1 = \dfrac{\cos^2\big(\frac{A}{2}\big)}{\sin^2\big({\frac{A}{2}}\big)} [/tex]
Next, we want to take advantage of the most essential trig identity, [tex] \sin^2{x}+\cos^2{x}=1 [/tex], specifically the form we get by subtracting [tex] \sin^2x [/tex] from either side: [tex] \cos^2{x}=1-\sin^2{x} [/tex]. We can transform the left side of this problem's equation into the same form by multiplying either side by [tex] \sin^2\big(\frac{A}{2}\big) [/tex]. Here's what we get out of that:
[tex] \sin^2\big(\frac{A}{2}\big)\Bigg(\dfrac{1}{\sin^2\big(\frac{A}{2}\big)} -1\Bigg) = \sin^2\big(\frac{A}{2}\big)\Bigg( \dfrac{\cos^2\big(\frac{A}{2}\big)}{\sin^2\big({\frac{A}{2}}\big)}\Bigg) \\\\\\ 1-\sin^2\big(\frac{A}{2}\big)=\cos^2\big(\frac{A}{2}\big) [/tex]
The identity [tex] \cos^2x=1-\sin^2x [/tex] tells us this statement must be true, so we're done at this point.
