Bentuk berikut yg ekuivalen dg sin x / 1+ cos x + 1+cos x /sin x adlah​

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Bentuk berikut yg ekuivalen dg sin x / 1+ cos x + 1+cos x /sin x adlah​

Jawab:

Penjelasan dengan langkah-langkah:

trigonometri

__

soal

sf dfrac{sin ~ x}{1 + cos~ x} + dfrac{1+ cos ~ x}{sin ~ x}=sf\\\sf dfrac{sin ~ x(sin ~ x) + (1 +cos~ x)(1 + cos~ x)}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{sin^2 ~ x + 1 +cos^2~ x+ 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{(sin^2 ~ x +cos^2~ x)+1 + 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{1+1 + 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{2 + 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\

sf dfrac{2 (1 + cos ~ x)}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{2}{sin~ x} = 2(dfrac{1}{sin~ x})= 2~cosec ~ x

Penjelasan dengan langkah-langkah:

trigonometri

__

soal

begin{gathered}sf dfrac{sin ~ x}{1 + cos~ x} + dfrac{1+ cos ~ x}{sin ~ x}=sf\\\sf dfrac{sin ~ x(sin ~ x) + (1 +cos~ x)(1 + cos~ x)}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{sin^2 ~ x + 1 +cos^2~ x+ 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{(sin^2 ~ x +cos^2~ x)+1 + 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{1+1 + 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{2 + 2 cos ~ x}{sin ~ x(1 + cos~ x)} sf\\end{gathered}

1+cos x

sin x

+

sin x

1+cos x

=

sin x(1+cos x)

sin x(sin x)+(1+cos x)(1+cos x)

sin x(1+cos x)

sin

2

x+1+cos

2

x+2cos x

sin x(1+cos x)

(sin

2

x+cos

2

x)+1+2cos x

sin x(1+cos x)

1+1+2cos x

sin x(1+cos x)

2+2cos x

begin{gathered}sf dfrac{2 (1 + cos ~ x)}{sin ~ x(1 + cos~ x)} sf\\\sf dfrac{2}{sin~ x} = 2(dfrac{1}{sin~ x})= 2~cosec ~ xend{gathered}

sin x(1+cos x)

2(1+cos x)

sin x

2

=2(

sin x

1

)=2 cosec x