Is 1+cot^2(x)=(tan^2(x)+1)/(tan^2(x)) ?

The answer to whether 1+cot^2(x)=(tan^2(x)+1)/(tan^2(x)) is True

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Popular Trigonometry >

prove 1+cot^2(x)=(tan^2(x)+1)/(tan^2(x))

Solution

prove

1 + cot^{2} (x) = \frac{tan ^{2} ( x ) + 1}{tan ^{2} ( x )}

Solution

True

Solution steps

1 + cot^{2} (x) = \frac{tan ^{2} ( x ) + 1}{tan ^{2} ( x )}

Manipulating left side

1 + cot^{2} (x)

Express with sin, cos

1 + cot^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

1 = \frac{1 sin ^{2} ( x )}{sin ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 1+cot^2(x)=(tan^2(x)+1)/(tan^2(x)) ?
The answer to whether 1+cot^2(x)=(tan^2(x)+1)/(tan^2(x)) is True