Is (sec^2(x)-1)csc(x)=sin(x)sec^2(x) ?

The answer to whether (sec^2(x)-1)csc(x)=sin(x)sec^2(x) is True

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Popular Trigonometry >

prove (sec^2(x)-1)csc(x)=sin(x)sec^2(x)

Solution

prove

(sec^{2} (x) - 1) csc (x) = sin (x) sec^{2} (x)

Solution

True

Solution steps

(sec^{2} (x) - 1) csc (x) = sin (x) sec^{2} (x)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(- a) = - a

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

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

(\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 )}

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

Join

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Cancel the common factor:

sin (x)

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{sin ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

Simplify

\frac{sin ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec^{2} (x) sin (x)

sec^{2} (x) sin (x)

sec^{2} (x) sin (x)

= sin (x) sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sec^2(x)-1)csc(x)=sin(x)sec^2(x) ?
The answer to whether (sec^2(x)-1)csc(x)=sin(x)sec^2(x) is True