Pembahasan Soal Trigonometri ~ Ruang Belajar Matematika

Kamis, 12 September 2019

Pembahasan Soal Trigonometri

Soal:
Jika   $\sec \alpha -\tan \alpha = x$ untuk $0^{o}< \alpha < 90^{o}$ , tentukanlah $\left ( \tan \alpha +\sec \alpha \right )$

Penyelesaian:
$\sec \alpha -\tan \alpha = x$
$(\sec \alpha - \tan \alpha )^{2} = x^{2}$
Uraian 1: $(\sec \alpha - \tan \alpha )^{2} = x^{2}$
   $\sec ^{2}\alpha -2.\sec \alpha.\tan \alpha + \tan ^{2}\alpha =x^{2}$
$\tan \alpha +1-2.\sec \alpha .\tan \alpha +\tan ^{2}\alpha =x^{2}$
$2.\tan ^{2}\alpha -2.\sec \alpha .\tan \alpha =x^{2}-1$
$2.\tan \alpha (\tan \alpha -\sec \alpha )=x^{2}-1$
$-2.\tan \alpha (\sec \alpha -\tan \alpha )=x^{2}-1$
   $-2.\tan \alpha .x=x^{2}-1$ ( lihat ketentuan di atas )

$\tan \alpha =\frac{x^{2}-1}{-2x}=-\frac{(x^{2}-1)}{2x}$

Uraian 2: $(\sec \alpha - \tan \alpha )^{2} = x^{2}$
$\sec ^{2}\alpha -2.\sec \alpha .\tan \alpha +\tan ^{2}\alpha =x^{2}$
   $\sec ^{2}\alpha -2.\sec \alpha .\tan \alpha +\sec ^{2}\alpha -1=x^{2}$
   $2.\sec ^{2}\alpha -2.\sec \alpha .\tan \alpha -1=x^{2}$
   $2.\sec \alpha (\sec \alpha -\tan \alpha )-1=x^{2}$
   $2.\sec \alpha .x=x^{2}+1$
   $\sec \alpha =\frac{(x^{2}+1)}{2x}$

$Jadi\, (\tan \alpha +\sec \alpha )=-\frac{(x^{2}-1)}{2x}+\frac{(x^{2}+1)}{2x}$
$=\frac{-x^{2}+1+x^{2}+1}{2x}$
$=\frac{2}{2x}=\frac{1}{x}$