apabila dik sinus b 12/13 tentukan nilai tangen b + cotangen b

Pembahasan.

Sin (b) = 12/13
Sin (b) = y/r

Kita cari x dengan teorema phytagoras

x² = r² - y²
x² = 13² - 12²
x² = 169 - 144
x² = 25
x = ± 5

Kuadran 1 ambil yang positif

Tan (b) = y/x
Tan (b) = 12/5

Cotan (b) = x/y
Cotan (b) = 5/12

Tan (b) + Cotan (b)
12/5 + 5/12
169/60 atau 2,81...

[tex]Sin \ b = \frac{y}{r}= \frac{12}{13} \\ y = 12, r = 13 \\ x= \sqrt{13^{2}-12^{2} }= \sqrt{169-144}= \sqrt{25} = 5 [/tex]

[tex]tan \ b = \frac{y}{x}= \frac{12}{5} \\ \\ cot \ b = \frac{x}{y} = \frac{5}{12} [/tex]

[tex]tan \ b + cot \ b = \frac{12}{5}+ \frac{5}{12} \\ \\ = \frac{169}{60} = 2,81[/tex]