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

sin (θ - \frac{π}{2}) = cos (π + θ)

\frac{1}{cot ( θ )} = \frac{sec ^{2} ( θ ) - 1}{tan ( θ )}

\frac{cos ( x )}{sin ( x ) + 1} = sec (x) - tan (x)

tan (- 7) = tan (π - 7)

sin (x) - 1 = 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