Is sec(2θ)=(csc^2(θ))/(csc^2(θ)-2) ?

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

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Popular >

prove sec(2θ)=(csc^2(θ))/(csc^2(θ)-2)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

Multiply

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

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

= \frac{1}{- 2 sin ^{2} ( θ ) + 1}

= \frac{1}{- 2 sin ^{2} ( θ ) + 1}

= \frac{1}{1 - 2 sin ^{2} ( θ )}

Rewrite using trig identities

\frac{1}{1 - 2 sin ^{2} ( θ )}

Use the Double Angle identity:

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

= \frac{1}{cos ( 2 θ )}

= \frac{1}{cos ( 2 θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{1}{\frac{1}{s e c ( 2 θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( 2 θ )}}

Apply the fraction rule:

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

= \frac{sec ( 2 θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (2 θ)

sec (2 θ)

sec (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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