Is (1+sin(α))/(cos(α))=(cos(α))/(1-sin(α)) ?

The answer to whether (1+sin(α))/(cos(α))=(cos(α))/(1-sin(α)) is True

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prove (1+sin(α))/(cos(α))=(cos(α))/(1-sin(α))

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 + sin ( α )}{cos ( α )}

Rewrite using trig identities

Multiply by

\frac{1 - sin ( α )}{1 - sin ( α )}

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

Expand

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

(1 + sin (α)) (1 - sin (α))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (α)

= 1^{2} - sin^{2} (α)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (α)

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

Use the Pythagorean identity:

1 = cos^{2} (α) + sin^{2} (α)

1 - sin^{2} (α) = cos^{2} (α)

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

Cancel the common factor:

cos (α)

= \frac{cos ( α )}{1 - sin ( α )}

= \frac{cos ( α )}{1 - sin ( α )}

We showed that the two sides could take the same form

\Rightarrow True

2 csc^{2} (y) = \frac{1}{1 - cos ( y )} + \frac{1}{1 + cos ( y )}

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

(tan (b) + sec (b)) (1 - sin (b)) = cos (b)

cot (2 x) = \frac{cos ( 2 x )}{sin ( 2 x )}

cos^{2} (a) + cos^{2} (a) cot^{2} (a) = cot^{2} (a)

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