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

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{csc ^{2} ( β )}

Express with sin, cos

\frac{1}{csc ^{2} ( β )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{sin ( β )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( β )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( β )}

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

Apply the fraction rule:

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

= \frac{sin ^{2} ( β )}{1}

Apply rule

\frac{a}{1} = a

= sin^{2} (β)

= sin^{2} (β)

= sin^{2} (β)

Rewrite using trig identities

sin^{2} (β)

Use the Pythagorean identity:

cos^{2} (β) + sin^{2} (β) = 1

sin^{2} (β) = 1 - cos^{2} (β)

= 1 - cos^{2} (β)

Factor

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

1 - cos^{2} (β)

Apply Difference of Two Squares Formula:

β^{2} - y^{2} = (β + y) (β - y)

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

= (1 + cos (β)) (1 - cos (β))

= (1 + cos (β)) (1 - cos (β))

= (1 + cos (β)) (1 - cos (β))

= (1 - cos (β)) (1 + cos (β))

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ( θ ) sec ( θ )}{csc ( θ )} = 1

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

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

\frac{1}{1 + tan ^{2} ( x )} = cos^{2} (x)

\frac{2 tan ( x )}{sec ^{2} ( x )} = sin (2 x)

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