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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

p ro v e 2 cos (2 x) = 4 cos^{2} (x) - 2

p ro v e tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

p ro v e \frac{1 - sin ( 2 x )}{cos ( 2 x )} = \frac{cos ( 2 x )}{1 + sin ( 2 x )}

p ro v e (csc (α) + cot (α)) (sec (α) - 1) = tan (α)

p ro v e 1 - 2 sin^{2} (θ) = 2 cos^{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