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人気のある三角関数 >

証明する (1-tan(θ))/(1+tan(θ))=(cot(θ)-1)/(cot(θ)+1)

解

証明する

\frac{1 - tan ( θ )}{1 + tan ( θ )} = \frac{cot ( θ ) - 1}{cot ( θ ) + 1}

解

真

解答ステップ

\frac{1 - tan ( θ )}{1 + tan ( θ )} = \frac{cot ( θ ) - 1}{cot ( θ ) + 1}

左側を操作する

\frac{1 - tan ( θ )}{1 + tan ( θ )}

サイン, コサインで表わす

\frac{1 - tan ( θ )}{1 + tan ( θ )}

基本的な三角関数の公式を使用する:

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

= \frac{1 - \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}}

簡素化

\frac{1 - \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}} : \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{1 - \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}}

結合

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

1 + \frac{sin ( θ )}{cos ( θ )}

元を分数に変換する:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

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 cos ( θ ) + sin ( θ )}{cos ( θ )}

乗算:

1 \cdot cos (θ) = cos (θ)

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

= \frac{1 - \frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

結合

1 - \frac{sin ( θ )}{cos ( θ )} : \frac{cos ( θ ) - sin ( θ )}{cos ( θ )}

1 - \frac{sin ( θ )}{cos ( θ )}

元を分数に変換する:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 \cdot cos ( θ )}{cos ( θ )} - \frac{sin ( θ )}{cos ( θ )}

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 cos ( θ ) - sin ( θ )}{cos ( θ )}

乗算:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) - sin ( θ )}{cos ( θ )}

= \frac{\frac{c o s ( θ ) - s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

分数を割る:

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

= \frac{( cos ( θ ) - sin ( θ ) ) cos ( θ )}{cos ( θ ) ( cos ( θ ) + sin ( θ ) )}

共通因数を約分する:

cos (θ)

= \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

= \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

= \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

右側を操作する

\frac{cot ( θ ) - 1}{cot ( θ ) + 1}

サイン, コサインで表わす

\frac{- 1 + cot ( θ )}{1 + cot ( θ )}

基本的な三角関数の公式を使用する:

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

= \frac{- 1 + \frac{c o s ( θ )}{s i n ( θ )}}{1 + \frac{c o s ( θ )}{s i n ( θ )}}

簡素化

\frac{- 1 + \frac{c o s ( θ )}{s i n ( θ )}}{1 + \frac{c o s ( θ )}{s i n ( θ )}} : \frac{- sin ( θ ) + cos ( θ )}{sin ( θ ) + cos ( θ )}

\frac{- 1 + \frac{c o s ( θ )}{s i n ( θ )}}{1 + \frac{c o s ( θ )}{s i n ( θ )}}

結合

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

1 + \frac{cos ( θ )}{sin ( θ )}

元を分数に変換する:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

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 sin ( θ ) + cos ( θ )}{sin ( θ )}

乗算:

1 \cdot sin (θ) = sin (θ)

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

= \frac{- 1 + \frac{c o s ( θ )}{s i n ( θ )}}{\frac{s i n ( θ ) + c o s ( θ )}{s i n ( θ )}}

結合

- 1 + \frac{cos ( θ )}{sin ( θ )} : \frac{- sin ( θ ) + cos ( θ )}{sin ( θ )}

- 1 + \frac{cos ( θ )}{sin ( θ )}

元を分数に変換する:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

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 sin ( θ ) + cos ( θ )}{sin ( θ )}

乗算:

1 \cdot sin (θ) = sin (θ)

= \frac{- sin ( θ ) + cos ( θ )}{sin ( θ )}

= \frac{\frac{- s i n ( θ ) + c o s ( θ )}{s i n ( θ )}}{\frac{s i n ( θ ) + c o s ( θ )}{s i n ( θ )}}

分数を割る:

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

= \frac{( - sin ( θ ) + cos ( θ ) ) sin ( θ )}{sin ( θ ) ( sin ( θ ) + cos ( θ ) )}

共通因数を約分する:

sin (θ)

= \frac{- sin ( θ ) + cos ( θ )}{sin ( θ ) + cos ( θ )}

= \frac{- sin ( θ ) + cos ( θ )}{sin ( θ ) + cos ( θ )}

= \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

証明する cot(θ)sin(θ)cos(θ)=1-sin^2(θ)

p ro v e cot (θ) sin (θ) cos (θ) = 1 - sin^{2} (θ)

証明する csc(θ)= 1/(sin(θ))

p ro v e csc (θ) = \frac{1}{sin ( θ )}

証明する 1/(sin(θ))-sin(θ)=cot(θ)cos(θ)

p ro v e \frac{1}{sin ( θ )} - sin (θ) = cot (θ) cos (θ)

証明する (sec(-x))/(tan(x)+cot(x))=-sin(-x)

p ro v e \frac{sec ( - x )}{tan ( x ) + cot ( x )} = - sin (- x)

証明する (sin(t))/(1-cos(t))=csc(t)+cot(t)

p ro v e \frac{sin ( t )}{1 - cos ( t )} = csc (t) + cot (t)