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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1 + \frac{1}{s i n ( x )}}{\frac{1}{s i n ( x )}}

Apply the fraction rule:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= sin (x) + 1

= sin (x) + 1

= 1 + sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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