Is sec(x)-sec(x)sin^2(x)=cos(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sec(x)-sec(x)sin^2(x)=cos(x)

prove

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

True

Solution steps

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

Manipulating left side

sec (x) - sec (x) sin^{2} (x)

Express with sin, cos

sec (x) - sec (x) sin^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{cos ( x )} sin^{2} (x)

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( x )}{cos ( x )}

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (2 x) = 2 sin (x) cos (x)

\frac{csc ^{2} ( x )}{cot ( x )} = csc (x) sec (x)

sec (x) + tan (x) = \frac{cos ( x )}{1 - sin ( x )}

csc (x) - cot (x) cos (x) = sin (x)

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

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