Is csc(θ)(1-cos(θ))(1+cos(θ))=sin(θ) ?

The answer to whether csc(θ)(1-cos(θ))(1+cos(θ))=sin(θ) is True

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prove csc(θ)(1-cos(θ))(1+cos(θ))=sin(θ)

prove

csc (θ) (1 - cos (θ)) (1 + cos (θ)) = sin (θ)

True

Solution steps

csc (θ) (1 - cos (θ)) (1 + cos (θ)) = sin (θ)

Manipulating left side

csc (θ) (1 - cos (θ)) (1 + cos (θ))

Rewrite using trig identities

csc (θ) (1 - cos (θ)) (1 + cos (θ))

Expand

(1 + cos (θ)) (1 - cos (θ)) : 1 - cos^{2} (θ)

(1 + cos (θ)) (1 - cos (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

= csc (θ) (1 - cos^{2} (θ))

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

= csc (θ) sin^{2} (θ)

= csc (θ) sin^{2} (θ)

Express with sin, cos

csc (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} sin^{2} (θ)

Simplify

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

\frac{1}{sin ( θ )} sin^{2} (θ)

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

csc^{2} (θ) + sec^{2} (θ) = csc^{2} (θ) sec^{2} (θ)

sin^{2} (x) = tan (x) cot (x) - cos^{2} (x)

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

cot^{4} (x) + 2 cot^{2} (x) = csc^{4} (x) - 1

Is csc(θ)(1-cos(θ))(1+cos(θ))=sin(θ) ?
The answer to whether csc(θ)(1-cos(θ))(1+cos(θ))=sin(θ) is True