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

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

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prove cos(-x)*cot(x)+sin(x)=csc(x)

prove

cos (- x) \cdot cot (x) + sin (x) = csc (x)

True

Solution steps

cos (- x) cot (x) + sin (x) = csc (x)

Manipulating left side

cos (- x) cot (x) + sin (x)

Use the negative angle identity:

cos (- x) = cos (x)

= sin (x) + cos (x) cot (x)

Express with sin, cos

sin (x) + cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{1}{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 )}}

Simplify

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= csc (x)

csc (x)

csc (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{0}{sin ( θ )} = \frac{1}{sin ( θ )} + 3

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

2 sin (x) - cot (x) = - cot (x) \cdot 2 cot (x)

4 cos^{3} (a) - 3 cos (a) = cos (3 a)

sin^{2} (θ) + (\frac{1}{2})^{2} = 1

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