Is cos(θ)sec(θ)-cos^2(θ)=sin^2(θ) ?

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

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prove cos(θ)sec(θ)-cos^2(θ)=sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

cos (θ) sec (θ) - cos^{2} (θ)

Express with sin, cos

- cos^{2} (θ) + cos (θ) sec (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

cos (θ) \frac{1}{cos ( θ )} = 1

cos (θ) \frac{1}{cos ( θ )}

Multiply fractions:

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

= \frac{1 \cdot cos ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= 1

= - 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

sin (12 x) = 2 sin (6 x) cos (6 x)

\frac{sin ( t )}{csc ( t )} = 1 - \frac{cos ( t )}{sec ( t )}

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

tan^{4} (θ) = tan^{2} (θ) sec^{2} (θ) - tan^{2} (θ)

\frac{sin ( 6 A ) + sin ( 2 A )}{cos ( 6 A ) - cos ( 2 A )} = - cot (2 A)

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