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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (a)

= cos (a)

= cos (a)

We showed that the two sides could take the same form

\Rightarrow True

\frac{3}{sec ( x ) - tan ( x )} = 3 (sec (x) + tan (x))

\frac{( cos ^{2} ( x ) + 4 cos ( x ) + 4 )}{( cos ( x ) + 2 )} = \frac{( 2 sec ( x ) + 1 )}{sec ( x )}

5 cos^{2} (x) - 5 sin^{2} (x) = 10 cos^{2} (x) - 5

sin (a) tan (a) + cos (a) = sec (a)

sin (x) (sec (x) - csc (x)) = tan (x) - 1

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