Question

7. Jika A dan B adalah sudut lancip dengan sin (A + B) = sqrt(5)/5 dan tan (À – B) = 1/3, Nilai dari tan A = ...
A. -1+√2
B. 1 + √2
C. 1+√3
D. 1-√3
E. √3 - 1 ​

Answer 1

Jawab: (A)

[tex]\sin A=\frac{x}{y}\rightarrow \cos A=\frac{\sqrt{y^2-x^2}}{y}\\\\\tan A=\frac{x}{y}\rightarrow \sin A=\dfrac{x}{\sqrt{x^2+y^2}}, \cos A=\dfrac{y}{\sqrt{x^2+y^2}}\\\sin (A+B)=\dfrac{\sqrt{5}}{5}=\dfrac{1}{\sqrt{5}}\rightarrow \cos(A+B)=\dfrac{2}{\sqrt{5}}\\\tan(A-B)=\dfrac{1}{3}\rightarrow \sin(A-B)=\dfrac{1}{\sqrt{1^2+3^2}}=\dfrac{1}{\sqrt{10}}, \cos(A-B)=\dfrac{3}{\sqrt{10}}[/tex]

[tex]\cos A\cos B = \frac{1}{2}(\cos (A+B)+\cos(A-B))\\\tan A =\dfrac{1}{2}(\dfrac{\sin(A+B)+\sin(A-B)}{\cos A\cos B})[/tex]

[tex]\cos A\cos B=\dfrac{1}{2}(\dfrac{2}{\sqrt{5}}+\dfrac{3}{\sqrt{10}}) \\\tan A=\dfrac{1}{2}(\dfrac{\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{10}}}{\dfrac{1}{2}(\dfrac{2}{\sqrt{5}}+\dfrac{3}{\sqrt{10}})})\times \dfrac{\sqrt{10}}{\sqrt{10}}= \dfrac{\sqrt{2}+1}{2\sqrt{2}+3}\times\dfrac{2\sqrt{2}-3}{2\sqrt{2}-3}\\\tan A=\dfrac{4-3\sqrt{2}+2\sqrt{2}-3}{8-9}=\dfrac{1-\sqrt{2}}{-1}=\sqrt{2}-1[/tex]