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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec (t) + tan (t)

Express with sin, cos

sec (t) + tan (t)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( t )} + tan (t)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{cos ( t )} + \frac{sin ( t )}{cos ( t )} : \frac{1 + sin ( t )}{cos ( t )}

\frac{1}{cos ( t )} + \frac{sin ( t )}{cos ( t )}

Apply rule

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

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

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

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

Rewrite using trig identities

Multiply by

\frac{1 - sin ( t )}{1 - sin ( t )}

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

Expand

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

(1 + sin (t)) (1 - sin (t))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (t)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (t)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (t)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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