quick inverse trig question (1 Viewer)

[kooltrainer]

show:
(tan inverse 2/root3 ) - ( tan inverse 3/2root3) = tan inverse root3 /12

 

[aalex]

Have fun!:wave:

 

[Trebla]

Let A = tan^-1(2 / √3) and B = tan^-1(3 / 2√3)
=> tan A = 2 / √3
=> tan B = 3 / 2√3

We know that:
tan (A - B) = [tan A - tan B] / [1 + tan A.tan B]
= [2 / √3 - 3 / 2√3] / [1 + 6/6]
= [1 / 2√3] / 2
= 1 / 4√3
= √3 / 12 after rationalising denominator
.: (A - B) = tan^-1(√3 / 12)
Sub back A and B:
.: tan^-1(2 / √3) - tan^-1(3 / 2√3) = tan^-1(√3 / 12)

 

[kooltrainer]

niiiceee...