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

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

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prove (csc^2(x)-1)/(cot(x))= 1/(tan(x))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{csc ^{2} ( x ) - 1}{cot ( x )}

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{cot ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Join

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

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

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

Manipulating right side

\frac{1}{tan ( x )}

Express with sin, cos

\frac{1}{tan ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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