Happy to help Gary.
Both in quadrant I makes it, if not easy, makes it definite. We don't have to consider multiple possibilities.
First let's write our goal, which is to fill out the tangent difference angle formula, which I'm geeky enough not to have to look up,
#tan(a-b) = {tan a - tan b}/{1 + tan a tan b}#
So we need #tan a# and #tan b# and we have sine or cosine. We could calculate the other sine or cosine and divide to get the tangent. But let's get the expressions symbolically first:
If #a#'s opposite is #p# and hypotenuse is #q# then # sin a = p/q # which we can write #a = arcsin(p/q)#. So the adjacent is #sqrt{q^2-p^2}# and the tangent is
# tan arcsin (p/q) = pm p/sqrt{q^2-p^2}#
Similarly,
#tan arccos(p/q) = pm sqrt{q^2-p^2}/p#
In this problem for #tan a# and #tan b# we can take the positive square root because we know they're in the first quadrant.
We note #7^2+24^2+25^2# is a Pythagorean Triple, as is #20^2+21^2=29^2#. I've done a lot of this stuff and I didn't recognize the second one.
#tan a = tan arcsin( 24/25) = 24/sqrt{25^2-24^2} = 24/7 #
#tan b = tan arccos( 20/29) = sqrt{29^2-20^2}/24 = 21/20 #
#tan(a-b) = {24/7 - 21/20}/{1 + 24/7 cdot 21/20} = 333/644
#
Check: Calculator
#tan(arcsin(24/25)-arccos(20/29)) times 644 = 333 quad sqrt#
