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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec (θ) - tan (θ)

Express with sin, cos

sec (θ) - tan (θ)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( θ )} - tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

= \frac{1 - sin ( θ )}{cos ( θ )}

= \frac{1 - sin ( θ )}{cos ( θ )}

= \frac{1 - sin ( θ )}{cos ( θ )}

Rewrite using trig identities

Multiply by

\frac{1 + sin ( θ )}{1 + sin ( θ )}

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

Expand

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

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

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

Cancel the common factor:

cos (θ)

= \frac{cos ( θ )}{1 + sin ( θ )}

= \frac{cos ( θ )}{1 + sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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sec^{4} (t) - tan^{4} (t) = 1 + 2 tan^{2} (t)

cos (4 θ) = cos^{2} (2 θ) - sin^{2} (2 θ)

sec (x) csc (x) - cot (x) = tan (x)

tan (θ) sin (2 θ) = 2 - 2 cos^{2} (θ)

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