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

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

prove

sec (x) - sec (x) \cdot 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

p ro v e \frac{2 sin ( x ) cos ( x )}{cos ( 2 x )} = tan (2 x)

p ro v e tan (- θ) = - tan (θ)

p ro v e tan (θ + π) = tan (θ)

p ro v e cos (\frac{π}{6}) = \frac{3}{2}

p ro v e sin (10 θ) = 2 sin (5 θ) cos (5 θ)

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