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

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{csc ^{2} ( B )}

Express with sin, cos

\frac{1}{csc ^{2} ( B )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{sin ( B )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (B)

= sin^{2} (B)

= sin^{2} (B)

Rewrite using trig identities

sin^{2} (B)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (B)

Factor

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

1 - cos^{2} (B)

Apply Difference of Two Squares Formula:

B^{2} - y^{2} = (B + y) (B - y)

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

= (1 + cos (B)) (1 - cos (B))

= (1 + cos (B)) (1 - cos (B))

= (1 + cos (B)) (1 - cos (B))

= (1 - cos (B)) (1 + cos (B))

We showed that the two sides could take the same form

\Rightarrow True

cot (θ) - tan (θ) = 2 cot (2 θ)

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

cot (2 x) = \frac{sec ( x ) + csc ( x )}{2 sin ( x ) - 2 cos ( x )}

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

\frac{\frac{1}{c s c ( x )}}{\frac{1}{c o t ( x )}} = cos (x)

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