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

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

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

prove

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

True

Solution steps

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

Manipulating right side

- tan (x) + sec (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

tan (a) + cot (a) = \frac{csc ( a )}{cos ( a )}

sin (x) tan (x) - csc (x) tan (x) = - cot (x)

sin (2 x) csc (2 x) = 1

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

sec^{2} (x) (1 + cos^{2} (x)) = 2 - tan^{2} (x)

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