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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec (β) + tan (β)

Express with sin, cos

sec (β) + tan (β)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( β )} + tan (β)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

= \frac{1 + sin ( β )}{cos ( β )}

= \frac{1 + sin ( β )}{cos ( β )}

= \frac{1 + sin ( β )}{cos ( β )}

Rewrite using trig identities

Multiply by

\frac{1 - sin ( β )}{1 - sin ( β )}

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

Expand

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

(1 + sin (β)) (1 - sin (β))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (β)

= 1^{2} - sin^{2} (β)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (β)

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

Use the Pythagorean identity:

1 = cos^{2} (β) + sin^{2} (β)

1 - sin^{2} (β) = cos^{2} (β)

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

Cancel the common factor:

cos (β)

= \frac{cos ( β )}{1 - sin ( β )}

= \frac{cos ( β )}{1 - sin ( β )}

We showed that the two sides could take the same form

\Rightarrow True

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

2 cot (2 θ) = cot (θ) - tan (θ)

7 cos^{2} (b) - 7 sin^{2} (b) = 14 cos^{2} (b) - 7

1 + tan (x) \cdot tan (2 x) = sec (2 x)

0 sec (0) = tan (0)

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