If 𝑚 is in Quadrant I and tan2𝑚 = 24/7 then the value of tan𝑚 is:
you have
y=24
x=7
so r = 25 (the familiar(!) 7-24-25 triangle)
That gives you
sin 2m = y/r = 24/25
cos 2m = x/r = 7/25
and now you can use the half-angle formula to find tan m
tan ( 2 θ ) = 2 tan θ / ( 1 - tan² θ )
tan ( 2 θ ) = 24 / 7
2 tan θ / ( 1 - tan² θ ) = 24 / 7
Divide both sides by 2.
tan θ / ( 1 - tan² θ ) = 12 / 7
Cross multiply.
tan θ • 7 = ( 1 - tan² θ ) • 12
7 tan θ = 12 - 12 tan² θ
Add 12 tan² θ to both sides.
12 tan² θ + 7 tan θ = 12
Subtract 12 to both sides.
12 tan² θ + 7 tan θ - 12 = 0
Mark tan θ as x
12 x² + 7 x - 12 = 0
The solutions are:
x = - 4 / 3 and x = 3 / 4
Since x = tan θ the solutions are:
tan θ = - 4 / 3 and tan θ = 3 / 4
In Quadrant I all trigonometric functions are positive, so:
tan θ = 3 / 4