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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec (u) + tan (u)

Express with sin, cos

sec (u) + tan (u)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

Multiply by

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

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

Expand

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

(1 + sin (u)) (1 - sin (u))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (u)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (u)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (u)

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos (x - \frac{3 π}{2}) = - sin (x)

\frac{sec ^{2} ( x )}{2 - sec ^{2} ( x )} = sec (2 x)

\frac{cot ( x ) tan ( x )}{cos ( x )} = sec (x)

\frac{sin ( θ )}{cos ( θ )} = tan (θ)

cos (x) tan (x) csc (x) = 1

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