Is sec(2θ)=(sec^2(θ))/(2-sec^2(θ)) ?

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

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Popular Trigonometry >

prove sec(2θ)=(sec^2(θ))/(2-sec^2(θ))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( θ )} ) ^{2}}{2 - ( \frac{1}{c o s ( θ )} ) ^{2}}

Simplify

\frac{( \frac{1}{c o s ( θ )} ) ^{2}}{2 - ( \frac{1}{c o s ( θ )} ) ^{2}} : \frac{1}{2 cos ^{2} ( θ ) - 1}

\frac{( \frac{1}{c o s ( θ )} ) ^{2}}{2 - ( \frac{1}{c o s ( θ )} ) ^{2}}

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{( \frac{1}{c o s ( θ )} ) ^{2}}{2 - \frac{1}{c o s ^{2} ( θ )}}

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{\frac{1}{c o s ^{2} ( θ )}}{2 - \frac{1}{c o s ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{1}{cos ^{2} ( θ ) ( 2 - \frac{1}{c o s ^{2} ( θ )} )}

Join

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

2 - \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{1}{\frac{2 c o s ^{2} ( θ ) - 1}{c o s ^{2} ( θ )} cos ^{2} ( θ )}

Multiply

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

cos^{2} (θ) \frac{2 cos ^{2} ( θ ) - 1}{cos ^{2} ( θ )}

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= 2 cos^{2} (θ) - 1

= \frac{1}{2 cos ^{2} ( θ ) - 1}

= \frac{1}{2 cos ^{2} ( θ ) - 1}

= \frac{1}{- 1 + 2 cos ^{2} ( θ )}

Rewrite using trig identities

\frac{1}{- 1 + 2 cos ^{2} ( θ )}

Use the Double Angle identity:

2 cos^{2} (x) - 1 = 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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Is sec(2θ)=(sec^2(θ))/(2-sec^2(θ)) ?
The answer to whether sec(2θ)=(sec^2(θ))/(2-sec^2(θ)) is True