Is cos^2(θ)(sec^2(θ)-1)=sin^2(θ) ?

The answer to whether cos^2(θ)(sec^2(θ)-1)=sin^2(θ) is True

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prove cos^2(θ)(sec^2(θ)-1)=sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (θ) (sec^{2} (θ) - 1)

Express with sin, cos

(- 1 + sec^{2} (θ)) cos^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )}

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

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{- cos ^{2} ( θ ) + 1}{cos ^{2} ( θ )} cos^{2} (θ)

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= - - cos^{2} (θ) + 1

= - cos^{2} (θ) + 1

= 1 - cos^{2} (θ)

Rewrite using trig identities

1 - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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