[tex](1 + \cos \alpha ) \csc \alpha + \frac{1}{ \csc \alpha(1 + \cos \alpha) } = 2 \csc \alpha [/tex]
Halo ada yg bisa bantu ngga? Terima kasih sebelumnya
TRigonomETRi
IdentitAs
Penjelasan dengan langkah-langkah:
buktikan . .
[tex]\sf (1 + \cos x) \csc x + \dfrac{1}{\csc x (1+ \cos\ x)} = 2 csc \ x[/tex]
ruas kiri
[tex]= \sf (1 + \cos x) \csc x + \dfrac{1}{\csc x (1+ \cos\ x)}[/tex]
[tex]= \sf (1 + \cos x) \frac{1}{sin \ x} + \dfrac{1}{\frac{1}{\sin x} (1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{(1 + \cos x) }{sin \ x} + \dfrac{\sin x}{(1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{(1 + \cos x)^2 + sin^2 x }{sin \ x(1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{(1 + 2\cos x + cos^2 \ x+ sin^2 x )}{sin \ x(1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{(1 + 2\cos x + 1)}{siin \ x(1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{(2+ 2\cos x )}{sin \ x(1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{2(1+ \cos x )}{sin \ x(1+ \cos\ x)}[/tex]
[tex]= \sf \dfrac{2 }{sin \ x} = 2\csc \ x[/tex]
terbukti sama ruas kanan
