Is sec^2(θ)=(sec^2(θ)-1)*csc^2(θ) ?

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

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prove sec^2(θ)=(sec^2(θ)-1)*csc^2(θ)

Solution

prove

sec^{2} (θ) = (sec^{2} (θ) - 1) \cdot csc^{2} (θ)

Solution

True

Solution steps

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

Manipulating right side

(sec^{2} (θ) - 1) csc^{2} (θ)

Express with sin, cos

(- 1 + sec^{2} (θ)) csc^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

Multiply fractions:

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

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

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

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

Multiply:

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

= (- 1 + \frac{1}{cos ^{2} ( θ )})

Remove parentheses:

(- a) = - a

= - 1 + \frac{1}{cos ^{2} ( θ )}

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

Join

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

- 1 + \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

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} ( θ ) + 1}{cos ^{2} ( θ )}

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin^{2} (θ)

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\frac{1}{sec ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sec ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{sec ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec^{2} (θ)

sec^{2} (θ)

sec^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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