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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

tan (x) (csc^{2} (x) - 1)

Express with sin, cos

(- 1 + csc^{2} (x)) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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