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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

cos (- x) = cos (x)

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

Rewrite using trig identities

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

Multiply by

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(1 - cos (x)) (1 + cos (x)) = 1 - cos^{2} (x)

= \frac{sin ( x )}{\frac{1 - c o s ^{2} ( x )}{1 + c o s ( x )}}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

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

= \frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

Simplify

\frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}} : \frac{1 + cos ( x )}{sin ( x )}

\frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

Apply the fraction rule:

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

= \frac{sin ( x ) ( 1 + cos ( x ) )}{sin ^{2} ( x )}

Cancel the common factor:

sin (x)

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

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

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

Manipulating right side

csc (x) + cot (x)

Express with sin, cos

cot (x) + csc (x)

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

tan (x) - tan (y) = \frac{sin ( x - y )}{cos ( x ) cos ( y )}

sec^{2} (x) tan^{2} (x) + sec^{2} (x) = sec^{4} (x)

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

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

1 + cos (2 x) + 2 sin^{2} (x) = 2

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