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

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

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

prove

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

True

Solution steps

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

Manipulating left side

- tan (y) + sec (y)

Express with sin, cos

sec (y) - tan (y)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( y )} - tan (y)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{cos ( y )} - \frac{sin ( y )}{cos ( y )} : \frac{1 - sin ( y )}{cos ( y )}

\frac{1}{cos ( y )} - \frac{sin ( y )}{cos ( y )}

Apply rule

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

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

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

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

Rewrite using trig identities

Multiply by

\frac{1 + sin ( y )}{1 + sin ( y )}

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

Expand

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

(1 + sin (y)) (1 - sin (y))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (y)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (y)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (y)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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