Is (1+sin(-x))/(1+csc(-x))=-sin(x) ?

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

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prove (1+sin(-x))/(1+csc(-x))=-sin(x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 + sin ( - x )}{1 + csc ( - x )}

Use the negative angle identity:

sin (- x) = - sin (x)

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

Use the negative angle identity:

csc (- x) = - csc (x)

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

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1 - sin ( x )}{1 - \frac{1}{s i n ( x )}}

Simplify

\frac{1 - sin ( x )}{1 - \frac{1}{s i n ( x )}} : - sin (x)

\frac{1 - sin ( x )}{1 - \frac{1}{s i n ( x )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

= \frac{1 - sin ( x )}{\frac{s i n ( x ) - 1}{s i n ( x )}}

Apply the fraction rule:

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

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

sin (x) - 1 = - (- sin (x) + 1)

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

Refine

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

Cancel the common factor:

1 - sin (x)

= - sin (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1 + csc ( x )}{sec ( x )} = cos (x) + cot (x)

cos (- x) = cos (x)

cos (π + x) = - cos (x)

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

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