Is sec(2x)=(sec^2(x))/(2-sec^2(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

prove sec(2x)=(sec^2(x))/(2-sec^2(x))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= \frac{1}{cos ( 2 x )}

= \frac{1}{cos ( 2 x )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( 2 x )}}

Simplify

\frac{1}{\frac{1}{s e c ( 2 x )}}

Apply the fraction rule:

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

= \frac{sec ( 2 x )}{1}

Apply rule

\frac{a}{1} = a

= sec (2 x)

sec (2 x)

sec (2 x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove tan(a)+cot(a)=sec(a)csc(a)

p ro v e tan (a) + cot (a) = sec (a) csc (a)

prove (1-sin(x))(1+sin(x))=cos^2(x)

p ro v e (1 - sin (x)) (1 + sin (x)) = cos^{2} (x)

prove (csc^2(x)-1)sin(x)=cos^2(x)csc(x)

p ro v e (csc^{2} (x) - 1) sin (x) = cos^{2} (x) csc (x)

prove sin(x)cot(x)=cos(x)

p ro v e sin (x) cot (x) = cos (x)

prove sin(pi+x)=-sin(x)

p ro v e sin (π + x) = - sin (x)

Frequently Asked Questions (FAQ)

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