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

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

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

prove

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

True

Solution steps

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

Manipulating right side

csc (x) - sin (x)

Express with sin, cos

csc (x) - sin (x)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{sin ( x )} - sin (x) : \frac{1 - sin ^{2} ( x )}{sin ( x )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (x) sin (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)

= 1 - sin^{2} (x)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( x )}{sin ( x )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (x) cot (x)

cos (x) cot (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (x + 27 0^{\circ}) = sin (x)

cos (u) sec (u) - sin^{2} (u) = cos^{2} (u)

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

cot (x) - 2 cot (2 x) = tan (x)

sin (x) = 3 sin (\frac{x}{2})

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