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

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

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prove (cos(2x))/(sin^2(x))=csc^2(x)-2

prove

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

True

Solution steps

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

Manipulating right side

csc^{2} (x) - 2

Express with sin, cos

- 2 + csc^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

- 2 + (\frac{1}{sin ( x )})^{2} : \frac{- 2 sin ^{2} ( x ) + 1}{sin ^{2} ( x )}

- 2 + (\frac{1}{sin ( 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 )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1 + csc ( a )}{cot ( a ) + cos ( a )} = sec (a)

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

- 2 = - 2 cos (0) + 15 sin (3) (0)

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

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