Is (1-cos(θ))(1+cos(θ))= 1/(csc^2(θ)) ?

The answer to whether (1-cos(θ))(1+cos(θ))= 1/(csc^2(θ)) is True

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prove (1-cos(θ))(1+cos(θ))= 1/(csc^2(θ))

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{csc ^{2} ( θ )}

Express with sin, cos

\frac{1}{csc ^{2} ( θ )}

Use the basic trigonometric identity:

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

= \frac{1}{( \frac{1}{s i n ( θ )} ) ^{2}}

Simplify

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

\frac{1}{( \frac{1}{s i n ( θ )} ) ^{2}}

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{\frac{1}{s i n ^{2} ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sin ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sin^{2} (θ)

= sin^{2} (θ)

= sin^{2} (θ)

Rewrite using trig identities

sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (θ) + sin^{2} (θ) = 1

sin^{2} (θ) = 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

Factor

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

1 - cos^{2} (θ)

Apply Difference of Two Squares Formula:

θ^{2} - y^{2} = (θ + y) (θ - y)

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

= (1 + cos (θ)) (1 - cos (θ))

= (1 + cos (θ)) (1 - cos (θ))

= (1 + cos (θ)) (1 - cos (θ))

= (1 - cos (θ)) (1 + cos (θ))

We showed that the two sides could take the same form

\Rightarrow True

sec (- x) sin (x) = tan (x)

cos (θ) csc (θ) tan (θ) = 1

cot^{2} (a) = cos^{2} (a) + (cot (a) cos (a))^{2}

tan^{4} (x) + 2 tan^{2} (x) + 1 = sec^{4} (x)

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

Is (1-cos(θ))(1+cos(θ))= 1/(csc^2(θ)) ?
The answer to whether (1-cos(θ))(1+cos(θ))= 1/(csc^2(θ)) is True