Is tan^3(x)cot(x)=sec^2(x)-1 ?

The answer to whether tan^3(x)cot(x)=sec^2(x)-1 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan^3(x)cot(x)=sec^2(x)-1

Solution

prove

tan^{3} (x) cot (x) = sec^{2} (x) - 1

Solution

True

Solution steps

tan^{3} (x) cot (x) = sec^{2} (x) - 1

Manipulating left side

tan^{3} (x) cot (x)

Express with sin, cos

cot (x) tan^{3} (x)

Use the basic trigonometric identity:

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

= \frac{cos ( x )}{sin ( x )} tan^{3} (x)

Use the basic trigonometric identity:

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

= \frac{cos ( x )}{sin ( x )} (\frac{sin ( x )}{cos ( x )})^{3}

Simplify

\frac{cos ( x )}{sin ( x )} (\frac{sin ( x )}{cos ( x )})^{3} : \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

\frac{cos ( x )}{sin ( x )} (\frac{sin ( x )}{cos ( x )})^{3}

Multiply fractions:

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

= \frac{cos ( x ) ( \frac{s i n ( x )}{c o s ( x )} ) ^{3}}{sin ( x )}

(\frac{sin ( x )}{cos ( x )})^{3} = \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

(\frac{sin ( x )}{cos ( x )})^{3}

Apply exponent rule:

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

= \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

= \frac{\frac{s i n ^{3} ( x )}{c o s ^{3} ( x )} cos ( x )}{sin ( x )}

Multiply

cos (x) \frac{sin ^{3} ( x )}{cos ^{3} ( x )} : \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

cos (x) \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

Multiply fractions:

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

= \frac{sin ^{3} ( x ) cos ( x )}{cos ^{3} ( x )}

Cancel the common factor:

cos (x)

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

= \frac{\frac{s i n ^{3} ( x )}{c o s ^{2} ( x )}}{sin ( x )}

Apply the fraction rule:

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

= \frac{sin ^{3} ( x )}{cos ^{2} ( x ) sin ( x )}

Cancel the common factor:

sin (x)

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

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

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

Manipulating right side

sec^{2} (x) - 1

Express with sin, cos

- 1 + sec^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove (sec^2(u)-1)/(sec^2(u))=sin^2(u)prove

\frac{sec ^{2} ( u ) - 1}{sec ^{2} ( u )} = sin^{2} (u)

prove sec(θ)=(tan(θ))/(csc(θ))+cos(θ)prove

sec (θ) = \frac{tan ( θ )}{csc ( θ )} + cos (θ)

prove sin((7pi)/6)=-1/2 prove

sin (\frac{7 π}{6}) = - \frac{1}{2}

prove cot(x)+tan(x)=(sec(x))/(sin(x))prove

cot (x) + tan (x) = \frac{sec ( x )}{sin ( x )}

prove 4cos^2(x)-4sin^2(x)=4-8sin^2(x)prove

4 cos^{2} (x) - 4 sin^{2} (x) = 4 - 8 sin^{2} (x)

Frequently Asked Questions (FAQ)

Is tan^3(x)cot(x)=sec^2(x)-1 ?
The answer to whether tan^3(x)cot(x)=sec^2(x)-1 is True