BANTU KAK PAKE CARANYAAAA

BANTU KAK PAKE CARANYAAAA

Jawab:

$bf b. bffrac{1}{3}left(2sqrt3-sqrt6right)$

Penjelasan dengan langkah-langkah:

$Largetext{$bffrac{tan52{,}5^{circ}-tan7{,}5^{circ}}{tan165^{circ}+tan75^{circ}}=...?$}$

1. Menyelesaikan tan 52,5° – tan 7,5°

Misalkan: tan 52,5° – tan 7,5° = tan A – tan B

$displaystylebegin{aligned}tan A-tan B&=tan(A-B)cdot(1+tan Atan B)\\&=frac{sin(A-B)}{cos(A-B)}cdotleft(1+frac{sin Asin B}{cos Acos B}right)\\&=frac{sin(A-B)}{cos(A-B)}cdotfrac{cos Acos B+sin Asin B}{cos Acos B}\\&=frac{sin(A-B)}{cos(A-B)}cdotfrac{cos(A-B)}{cos Acos B}\\&=largetext{$frac{sin(A-B)}{cos Acos B}$}end{aligned}$

Untuk cos A cos B:

$displaystylebegin{aligned}cos(A+B)&=cos Acos B-sin Asin B\cos(A-B)&=cos Acos B+sin Asin B +\cline{1-2}cos(A+B)+cos(A-B)&=2(cos Acos B)\end{aligned}\\\text{Maka:quad}boxed{largetext{$cos Acos B=frac{1}{2}big[cos(A+B)+cos(A-B)big]$}}$

Substitusi nilai cos A cos B:

$displaystylebegin{aligned}tan A-tan B&=frac{sin(A-B)}{cos Acos B}\\&=frac{sin(A-B)}{frac{1}{2}big[cos(A+B)+cos(A-B)big]}\\&=largetext{$frac{2sin(A-B)}{cos(A+B)+cos(A-B)}$}end{aligned}$

Oleh karena itu, tan 52,5° – tan 7,5° bisa dihitung sbb.:

$displaystylebegin{aligned}tan52{,}5^{circ}-tan7{,}5^{circ}&=frac{2sin(52{,}5^{circ}-7{,}5^{circ})}{cos(52{,}5^{circ}+7{,}5^{circ})+cos(52{,}5^{circ}-7{,}5^{circ})}\\&=frac{2sin{45^{circ}}}{cos{60^{circ}}+cos{45^{circ}}}\\&=frac{2cdotfrac{1}{2}sqrt2}{frac{1}{2}+frac{1}{2}sqrt2}=frac{sqrt2}{frac{1}{2}(1+sqrt2)}\\&=frac{2sqrt2}{1+sqrt2}cdotfrac{1-sqrt2}{1-sqrt2}\\&=frac{2sqrt2-4}{1-2}\\&=largetext{$4-2sqrt2$}end{aligned}$

2. Menyelesaikan tan 165°

$displaystylebegin{aligned}tan165^{circ}&=tan(120^{circ}+45^{circ})\\&=frac{tan120^{circ}+tan45^{circ}}{1-tan120^{circ}tan45^{circ}}\\&=frac{1-sqrt3}{1+sqrt3}cdotleft(frac{1-sqrt3}{1-sqrt3}right)\\&=frac{1-2sqrt3+3}{1-3}=frac{4-2sqrt3}{-2}\\&=largetext{$sqrt3-2$}end{aligned}$

3. Menyelesaikan tan 75°

$displaystylebegin{aligned}tan75^{circ}&=tan(45^{circ}+30^{circ})\\&=frac{tan45^{circ}+tan30^{circ}}{1-tan45^{circ}tan30^{circ}}\\&=frac{1+frac{1}{3}sqrt3}{1-frac{1}{3}sqrt3}cdotleft(frac{1+frac{1}{3}sqrt3}{1+frac{1}{3}sqrt3}right)\\&=frac{1+frac{2}{3}sqrt3+frac{1}{3}}{1-frac{1}{3}}=frac{3left(1+frac{2}{3}sqrt3+frac{1}{3}right)}{2}\\&=frac{3+2sqrt3+1}{2}\\&=largetext{$2+sqrt3$}end{aligned}$

4. Penyelesaian akhir

tan 52,5° – tan 7,5° = 4 – 2√2
tan 165° = √3 – 2
tan 75° = 2 + √3

$displaystylebegin{aligned}largetext{$bffrac{tan52{,}5^{circ}-tan7{,}5^{circ}}{tan165^{circ}+tan75^{circ}}$}&=frac{4-2sqrt2}{sqrt3-2+2+sqrt3}\\&=frac{4-2sqrt2}{2sqrt3}cdotleft(frac{sqrt3}{sqrt3}right)\\&=frac{4sqrt3-2sqrt6}{6}=frac{2sqrt3-sqrt6}{3}\\&=Largetext{$bffrac{1}{3}left(2sqrt3-sqrt6right)$}end{aligned}$