One approach is to start from the standard formula
[tex]\sin^2(x) + cos^2(x) = 1[/tex]
and manipulate the left side to be what we want and then try to clean up the right side after.
To get [tex]\tan[/tex] on the left side, we divide everything by [tex]\cos^2(x)[/tex]:
[tex]\dfrac{\sin^2(x)}{cos^2(x)} + \dfrac{cos^2(x)}{cos^2(x)} = \dfrac{1}{cos^2(x)}[/tex]
or
[tex]\tan^2(x)+1=\sec^2(x)[/tex]
Reordering the left side gives us
[tex]1+\tan^2(x)=\sec^2(x)[/tex]
Now this is true for all values of x, including the value [tex]\frac{a+b}{2}[/tex].
So we now have:
[tex]1+\tan^2(\frac{a+b}{2})=\sec^2(\frac{a+b}{2})[/tex]
and then left side of the equation is what we want.
Now to manipulate the right side...
We going to use several facts about this situation.
So here's how we'll use these facts.
1.
Since [tex]a+b+c=\pi[/tex], we can also say [tex]a+b=\pi-c[/tex] and also that [tex]\dfrac{a+b}{2}=\dfrac{\pi-c}{2}[/tex] or [tex]\dfrac{a+b}{2}=\dfrac{\pi}{2}-\dfrac{c}{2}[/tex]. This allows us to rewrite our equation as
[tex]1+\tan^2(\frac{a+b}{2})=\sec^2(\frac{\pi}{2}-\frac{c}{2})[/tex]
2.
Since [tex]\sec(x)[/tex] is even, [tex]\sec(\frac{\pi}{2}-\frac{c}{2})=\sec(-1\cdot(\frac{\pi}{2}-\frac{c}{2}))=\sec(\frac{c}{2}-\frac{\pi}{2})[/tex]. This allows us to further rewrite our equation as
[tex]1+\tan^2(\frac{a+b}{2})=\sec^2(\frac{c}{2}-\frac{\pi}{2})[/tex]
3.
Finally, since [tex]\csc(x)[/tex] same as [tex]\sec(x-\frac{\pi}{2})[/tex], we can say that [tex]\sec(\frac{c}{2}-\frac{\pi}{2})=\csc(\frac{c}{2})[/tex], which extends to [tex]\sec^2(\frac{c}{2}-\frac{\pi}{2})=\csc^2(\frac{c}{2})[/tex]. That gives us the final piece of the right side:
[tex]1+\tan^2(\frac{a+b}{2})=\sec^2(\frac{c}{2}-\frac{\pi}{2})[/tex]
is now
[tex]1+\tan^2(\frac{a+b}{2})=\csc^2(\frac{c}{2})[/tex]
