Is sec(θ)-sec(θ)sin^2(θ)=cos(θ) ?

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

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prove sec(θ)-sec(θ)sin^2(θ)=cos(θ)

prove

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

True

Solution steps

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

Manipulating left side

sec (θ) - sec (θ) sin^{2} (θ)

Express with sin, cos

sec (θ) - sec (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{cos ( θ )} sin^{2} (θ)

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

- 2 = - 2 cos (0) + 15 sin (3) (0)

tan (x) - sin (x) = tan (x) sin (x)

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

sin (x) + \frac{cos ( x )}{tan ( x )} = csc (x)

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

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