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人気のある三角関数 >

証明する csc(x)(sin(x)+tan(x))=1+sec(x)

解

証明する

csc (x) (sin (x) + tan (x)) = 1 + sec (x)

解

真

解答ステップ

csc (x) (sin (x) + tan (x)) = 1 + sec (x)

左側を操作する

csc (x) (sin (x) + tan (x))

サイン, コサインで表わす

(sin (x) + tan (x)) csc (x)

基本的な三角関数の公式を使用する:

tan (x) = \frac{sin ( x )}{cos ( x )}

= (sin (x) + \frac{sin ( x )}{cos ( x )}) csc (x)

基本的な三角関数の公式を使用する:

csc (x) = \frac{1}{sin ( x )}

= (sin (x) + \frac{sin ( x )}{cos ( x )}) \frac{1}{sin ( x )}

簡素化

(sin (x) + \frac{sin ( x )}{cos ( x )}) \frac{1}{sin ( x )} : \frac{cos ( x ) + 1}{cos ( x )}

(sin (x) + \frac{sin ( x )}{cos ( x )}) \frac{1}{sin ( x )}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( sin ( x ) + \frac{s i n ( x )}{c o s ( x )} )}{sin ( x )}

1 \cdot (sin (x) + \frac{sin ( x )}{cos ( x )}) = sin (x) + \frac{sin ( x )}{cos ( x )}

1 \cdot (sin (x) + \frac{sin ( x )}{cos ( x )})

乗算:

1 \cdot (sin (x) + \frac{sin ( x )}{cos ( x )}) = (sin (x) + \frac{sin ( x )}{cos ( x )})

= (sin (x) + \frac{sin ( x )}{cos ( x )})

括弧を削除する:

(a) = a

= sin (x) + \frac{sin ( x )}{cos ( x )}

= \frac{sin ( x ) + \frac{s i n ( x )}{c o s ( x )}}{sin ( x )}

結合

sin (x) + \frac{sin ( x )}{cos ( x )} : \frac{sin ( x ) cos ( x ) + sin ( x )}{cos ( x )}

sin (x) + \frac{sin ( x )}{cos ( x )}

元を分数に変換する:

sin (x) = \frac{sin ( x ) cos ( x )}{cos ( x )}

= \frac{sin ( x ) cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) cos ( x ) + sin ( x )}{cos ( x )}

= \frac{\frac{s i n ( x ) c o s ( x ) + s i n ( x )}{c o s ( x )}}{sin ( x )}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{sin ( x ) cos ( x ) + sin ( x )}{cos ( x ) sin ( x )}

共通項をくくり出す

sin (x)

= \frac{sin ( x ) ( cos ( x ) + 1 )}{cos ( x ) sin ( x )}

共通因数を約分する:

sin (x)

= \frac{cos ( x ) + 1}{cos ( x )}

= \frac{cos ( x ) + 1}{cos ( x )}

= \frac{1 + cos ( x )}{cos ( x )}

三角関数の公式を使用して書き換える

基本的な三角関数の公式を使用する:

cos (x) = \frac{1}{sec ( x )}

\frac{1 + \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

簡素化

\frac{1 + \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

分数の規則を適用する:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( 1 + \frac{1}{s e c ( x )} ) sec ( x )}{1}

結合

1 + \frac{1}{sec ( x )} : \frac{sec ( x ) + 1}{sec ( x )}

1 + \frac{1}{sec ( x )}

元を分数に変換する:

1 = \frac{1 sec ( x )}{sec ( x )}

= \frac{1 \cdot sec ( x )}{sec ( x )} + \frac{1}{sec ( x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot sec ( x ) + 1}{sec ( x )}

乗算:

1 \cdot sec (x) = sec (x)

= \frac{sec ( x ) + 1}{sec ( x )}

= \frac{\frac{s e c ( x ) + 1}{s e c ( x )} sec ( x )}{1}

分数の規則を適用する:

\frac{a}{1} = a

= \frac{sec ( x ) + 1}{sec ( x )} sec (x)

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( sec ( x ) + 1 ) sec ( x )}{sec ( x )}

共通因数を約分する:

sec (x)

= sec (x) + 1

sec (x) + 1

sec (x) + 1

= 1 + sec (x)

両辺を同じ形式にできることを証明した

\Rightarrow 真

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