Is sec(A)-tan(A)=(cos(A))/(1+sin(A)) ?

The answer to whether sec(A)-tan(A)=(cos(A))/(1+sin(A)) is True

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prove sec(A)-tan(A)=(cos(A))/(1+sin(A))

prove

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

True

Solution steps

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

Manipulating left side

sec (A) - tan (A)

Express with sin, cos

sec (A) - tan (A)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( A )} - tan (A)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( A )} - \frac{sin ( A )}{cos ( A )}

Simplify

\frac{1}{cos ( A )} - \frac{sin ( A )}{cos ( A )} : \frac{1 - sin ( A )}{cos ( A )}

\frac{1}{cos ( A )} - \frac{sin ( A )}{cos ( A )}

Apply rule

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

= \frac{1 - sin ( A )}{cos ( A )}

= \frac{1 - sin ( A )}{cos ( A )}

= \frac{1 - sin ( A )}{cos ( A )}

Rewrite using trig identities

Multiply by

\frac{1 + sin ( A )}{1 + sin ( A )}

= \frac{( 1 + sin ( A ) ) ( 1 - sin ( A ) )}{( 1 + sin ( A ) ) cos ( A )}

Expand

(1 + sin (A)) (1 - sin (A)) : 1 - sin^{2} (A)

(1 + sin (A)) (1 - sin (A))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (A)

= 1^{2} - sin^{2} (A)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (A)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (A)

= \frac{cos ( A )}{1 + sin ( A )}

= \frac{cos ( A )}{1 + sin ( A )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

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

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

tan^{3} (x) = (tan (x))^{3}

Is sec(A)-tan(A)=(cos(A))/(1+sin(A)) ?
The answer to whether sec(A)-tan(A)=(cos(A))/(1+sin(A)) is True