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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec (x) + tan (x)

Express with sin, cos

sec (x) + tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

Multiply by

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

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

Expand

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

(1 + sin (x)) (1 - sin (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

arccsc (- 2)

sin^{2} (x) = sin (x)

cos (22 5^{\circ})

arctan (\frac{5}{2}) \cdot \frac{180}{π}

cos (3 x) = \frac{1}{2}

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