Nilai dari (tan 75° + tan 15°)/(tan 75° - tan 15°) adalah...

Jawabnya adalah ⅔√3 Ingat tan (A+B) = ( tan A + tan B)/(1- (tan A tan B)) tan (A-B) = ( tan A - tan B)/(1+ (tan A tan B)) Sudut istimewa dalam identitas trigonometri tan 30° = ⅓√3 tan 45° = 1 (tan 75° + tan 15°)/(tan 75° - tan 15°) = (tan (45°+30°)+ tan (45°-30°)) / (tan (45°+30°) - tan (45°-30°)) = [((tan 45° + tan 30°)/(1- tan 45° tan 30°)) + ((tan 45° -tan 30°)/(1 + tan 45° tan 30°)) ] / [((tan 45° + tan 30°)/(1- tan 45° tan 30°)) - ((tan 45° -tan 30°)/(1 + tan 45° tan 30°)) ] = [((1 + ⅓√3)/(1- 1.⅓√3)) + ((1 -⅓√3)/(1 + 1. ⅓√3)) ] / [((1 +⅓√3)/(1- 1.⅓√3)) - ((1 -⅓√3)/(1 + 1.⅓√3)) ] = [((1 + ⅓√3)/(1- ⅓√3)) + ((1 -⅓√3)/(1 + ⅓√3)) ] / [((1 +⅓√3)/(1- ⅓√3)) - ((1 -⅓√3)/(1 + ⅓√3)) ] = [((1 + ⅓√3)(1+ ⅓√3)) + ((1 -⅓√3)(1 -⅓√3)) /((1+⅓√3)(1-⅓√3))] / [((1 + ⅓√3)(1+ ⅓√3)) -((1 -⅓√3)(1 -⅓√3)) /((1+⅓√3)(1-⅓√3))] = [((1 + ⅓√3)(1+ ⅓√3)) + ((1 -⅓√3)(1 -⅓√3)) ]/ [((1 + ⅓√3)(1+ ⅓√3)) -((1 -⅓√3)(1 -⅓√3))] = [(1²+⅔√3+(⅓√3)²) + (1²-⅔√3 +(⅓√3)²)]/ [(1²+⅔√3 +(⅓√3)²) - (1²-⅔√3 +(⅓√3)²)] = [(1+3/9 + ⅔√3)+(1+3/9-⅔√3)]/[(1+3/9+ ⅔√3) -(1+3/9-⅔√3)] = [2+3/9 + 3/9]/[(4√3)/3] = (2+6/9) /((4√3)/3) = (8/3).(3/(4√3)) = (8/(4√3)). (√3/√3) = (8√3)/(4.3) = (8√3)/12 = ⅔√3 Jadi Jawabnya adalah ⅔√3