Is (cos^2(x))/(sin(x))=csc(x)-sin(x) ?

The answer to whether (cos^2(x))/(sin(x))=csc(x)-sin(x) is True

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prove (cos^2(x))/(sin(x))=csc(x)-sin(x)

prove

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

True

Solution steps

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

We showed that the two sides could take the same form

\Rightarrow True

1 - 4 sin^{2} (x) cos^{2} (x) = cos^{2} (2 x)

sin (3 x) = sin (x) (4 cos^{2} (x) - 1)

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

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

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

Is (cos^2(x))/(sin(x))=csc(x)-sin(x) ?
The answer to whether (cos^2(x))/(sin(x))=csc(x)-sin(x) is True