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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

Rewrite using trig identities

sin^{2} (x)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (x)

Factor

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

1 - cos^{2} (x)

Apply Difference of Two Squares Formula:

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

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

= (1 + cos (x)) (1 - cos (x))

= (1 + cos (x)) (1 - cos (x))

= (1 + cos (x)) (1 - cos (x))

= (1 - cos (x)) (1 + cos (x))

We showed that the two sides could take the same form

\Rightarrow True

csc (x) cos (x) = cot (x)

sin (π - x) = sin (x)

\frac{sin ( 2 θ )}{1 - cos ( 2 θ )} = cot (θ)

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

cos (θ) + sin (θ) tan (θ) = sec (θ)

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