Is (csc^2(θ)-2)/(csc^2(θ))=cos(2θ) ?

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

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prove (csc^2(θ)-2)/(csc^2(θ))=cos(2θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ^{2} ( θ ) - 2}{csc ^{2} ( θ )}

Express with sin, cos

\frac{- 2 + csc ^{2} ( θ )}{csc ^{2} ( θ )}

Use the basic trigonometric identity:

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

= \frac{- 2 + ( \frac{1}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}}

Simplify

\frac{- 2 + ( \frac{1}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}} : - 2 sin^{2} (θ) + 1

\frac{- 2 + ( \frac{1}{s i n ( θ )} ) ^{2}}{( \frac{1}{s i n ( θ )} ) ^{2}}

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{- 2 + ( \frac{1}{s i n ( θ )} ) ^{2}}{\frac{1}{s i n ^{2} ( θ )}}

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{- 2 + \frac{1}{s i n ^{2} ( θ )}}{\frac{1}{s i n ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{( - 2 + \frac{1}{s i n ^{2} ( θ )} ) sin ^{2} ( θ )}{1}

Join

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

- 2 + \frac{1}{sin ^{2} ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{\frac{- 2 s i n ^{2} ( θ ) + 1}{s i n ^{2} ( θ )} sin ^{2} ( θ )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= - - 2 sin^{2} (θ) + 1

= - 2 sin^{2} (θ) + 1

= 1 - 2 sin^{2} (θ)

Rewrite using trig identities

1 - 2 sin^{2} (θ)

Use the Double Angle identity:

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

= cos (2 θ)

= cos (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos (b) + sin (b) tan (b) = sec (b)

2 sin (2 θ) (1 - 2 sin^{2} (θ)) = sin (4 θ)

cos^{2} (x) csc^{2} (x) = cot^{2} (x)

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