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

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

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

prove

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

True

Solution steps

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

Manipulating right side

csc (θ) - sin (θ)

Express with sin, cos

csc (θ) - sin (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} - sin (θ)

Simplify

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

\frac{1}{sin ( θ )} - sin (θ)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - sin (θ) sin (θ) = 1 - sin^{2} (θ)

1 - sin (θ) sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= 1 - sin^{2} (θ)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

7 sec (y) cos (y) = 7

\frac{1 + cos ( 2 t )}{sin ( 2 t ) - cos ( t )} = \frac{2 cos ( t )}{2 sin ( t ) - 1}

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