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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(- a) = - a

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos^{2} (x)

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc^{2} (x)

csc^{2} (x)

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