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

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

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

prove

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

True

Solution steps

\frac{1}{sec ( 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 )}

Use the basic trigonometric identity:

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

\frac{\frac{1}{s e c ( x )}}{1 - sin ( x )}

Simplify

\frac{\frac{1}{s e c ( x )}}{1 - sin ( x )}

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

- csc (x) cos (x) = - cot (x)

sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

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

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

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