Matematika Sekolah Menengah Atas 1.cos B cot B =csc B-sinB
2.sin²A (1+cot²A)=1



saya mohon bantu saya​

1.cos B cot B =csc B-sinB
2.sin²A (1+cot²A)=1



saya mohon bantu saya​

Penjelasan dengan langkah-langkah:

Nomor 1

[tex] = \cos(b) \cot(b) [/tex]

[tex] = \cos(b) . \frac{ \cos(b) }{ \sin(b) } [/tex]

[tex] = \frac{ \cos {}^{2} (b) }{ \sin(b) } [/tex]

[tex] = \frac{1 - \sin {}^{2} (b) }{ \sin(b) } [/tex]

[tex] = \frac{1}{ \sin(b) } - \frac{ \sin {}^{2} (b) }{ \sin(b) } [/tex]

[tex] = \csc(b) - \sin(b) \: \text{terbukti}[/tex]

[tex] \: [/tex]

Nomor 2

[tex] = \sin {}^{2} (a) .(1 + \cot {}^{2} (a) )[/tex]

[tex] = \sin {}^{2} (a) .( \frac{ \sin {}^{2} (a) }{ \sin {}^{2} (a) } + \frac{ \cos {}^{2} (a) }{ \sin {}^{2} (a) } )[/tex]

[tex] = \sin {}^{2} (a) .( \frac{ \sin {}^{2} (a) + \cos {}^{2} (a) }{ \sin {}^{2} (a) } )[/tex]

[tex] = \cancel{\sin {}^{2} (a) }.( \frac{ \sin {}^{2} (a) + \cos {}^{2} (a) }{ \cancel{\sin {}^{2} (a) } })[/tex]

[tex] = \sin {}^{2} (a) + \cos {}^{2} (a) [/tex]

[tex] = 1 \: \text{terbukti}[/tex]

[answer.2.content]