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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

(sec^{2} (θ) - 1) (sin^{2} (θ) - 1) = - sin^{2} (θ)

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

csc^{2} (x) (1 - cos^{2} (x)) = sin (648 0^{\circ})

- 2 cos^{4} (x) = - 2 (cos^{2} (x)) (cos^{2} (x))

csc (θ) = \frac{tan ( θ ) cot ( θ )}{sin ( θ )}

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