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

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

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

prove

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

True

Solution steps

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

Manipulating left 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

sin^{2} (x) - 3 cos^{2} (x) = 4 sin^{2} (x) - 3

tan^{2} (x) + 6 = sec^{2} (x) + 5

tan (\frac{π}{2} - θ) tan (θ) = 1

(1 - sin^{2} (x)) csc^{2} (x) = cot^{2} (x)

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

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