Is sin(θ)(csc(θ)-sin(θ))=cos^2(θ) ?

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

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prove sin(θ)(csc(θ)-sin(θ))=cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin (θ) (csc (θ) - sin (θ))

Express with sin, cos

(csc (θ) - sin (θ)) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

\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{- sin ^{2} ( θ ) + 1}{sin ( θ )} sin (θ)

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

Rewrite using trig identities

1 - sin^{2} (θ)

Use the Pythagorean identity:

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

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

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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