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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (x) (1 + sin (x))

Express with sin, cos

(1 + sin (x)) csc (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

1 \cdot (1 + sin (x))

Multiply:

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

= (1 + sin (x))

Remove parentheses:

(a) = a

= 1 + sin (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1 + \frac{1}{c s c ( x )}}{\frac{1}{c s c ( x )}}

Simplify

\frac{1 + \frac{1}{c s c ( x )}}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

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

= \frac{( 1 + \frac{1}{c s c ( x )} ) csc ( x )}{1}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot csc (x) = csc (x)

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

= \frac{\frac{c s c ( x ) + 1}{c s c ( x )} csc ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

csc (x)

= csc (x) + 1

csc (x) + 1

csc (x) + 1

= 1 + csc (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (x + y) sin (x - y) = sin (2 x) - sin (2 y)

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

cos^{2} (θ) = 1 + sin^{2} (θ)

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

sin (\frac{5 π}{6}) = \frac{1}{2}

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