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人気のある三角関数 >

証明する (tan^2(θ)+1)/(tan^2(θ))=csc^2(θ)

解

証明する

\frac{tan ^{2} ( θ ) + 1}{tan ^{2} ( θ )} = csc^{2} (θ)

解

真

解答ステップ

\frac{tan ^{2} ( θ ) + 1}{tan ^{2} ( θ )} = csc^{2} (θ)

左側を操作する

\frac{tan ^{2} ( θ ) + 1}{tan ^{2} ( θ )}

サイン, コサインで表わす

\frac{1 + tan ^{2} ( θ )}{tan ^{2} ( 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 )} ) ^{2}}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

簡素化

\frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}} : \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

\frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

(\frac{sin ( θ )}{cos ( θ )})^{2} = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

(\frac{sin ( θ )}{cos ( θ )})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

分数の規則を適用する:

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

= \frac{( 1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )} ) cos ^{2} ( θ )}{sin ^{2} ( θ )}

結合

1 + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} : \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

1 + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

元を分数に変換する:

1 = \frac{1 cos ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{1 \cdot cos ^{2} ( θ )}{cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{cos ^{2} ( 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 ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

乗算:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{\frac{c o s ^{2} ( θ ) + s i n ^{2} ( θ )}{c o s ^{2} ( θ )} cos ^{2} ( θ )}{sin ^{2} ( θ )}

乗じる

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} cos^{2} (θ) : cos^{2} (θ) + sin^{2} (θ)

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} cos^{2} (θ)

分数を乗じる:

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

= \frac{( cos ^{2} ( θ ) + sin ^{2} ( θ ) ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

共通因数を約分する:

cos^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

三角関数の公式を使用して書き換える

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

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:

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

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

簡素化

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( θ )}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= \frac{1}{\frac{1}{c s c ^{2} ( θ )}}

分数の規則を適用する:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( θ )}{1}

規則を適用

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{2} (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 sec^2(x)csc^2(x)-sec^2(x)=csc^2(x)

p ro v e sec^{2} (x) csc^{2} (x) - sec^{2} (x) = csc^{2} (x)

証明する (sin(θ)+sin(3θ))/(2cos(θ))=sin(2θ)

p ro v e \frac{sin ( θ ) + sin ( 3 θ )}{2 cos ( θ )} = sin (2 θ)

証明する csc(θ)-csc(θ)cos^2(θ)=sin(θ)

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

証明する cot(2β)=(cot^2(β)-1)/(2cot(β))

p ro v e cot (2 β) = \frac{cot ^{2} ( β ) - 1}{2 cot ( β )}

証明する (csc(x)-cot(x))/(csc(x))=1-cos(x)

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