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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

tan^{5} (x) = tan^{3} (x) sec^{2} (x) - tan^{3} (x)

\frac{cos ( x )}{cot ( x )} = sin (x)

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

cos (x) csc (x) = cot (x)

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