Is sec(2x)=(sec^2(x))/(1-tan^2(x)) ?

The answer to whether sec(2x)=(sec^2(x))/(1-tan^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))/(1-tan^2(x))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

\frac{sec ^{2} ( x )}{1 - tan ^{2} ( x )}

Express with sin, cos

\frac{sec ^{2} ( x )}{1 - tan ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = 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 (sin(3x)-sin(5x))/(2cos(4x))=-sin(x)

prove tan(a)+cot(a)=2csc(2a)

prove (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1)

prove 3cos^2(x)-3sin^2(x)=3-6sin^2(x)

prove sec^4(x)= 1/(cos^4(x))

Frequently Asked Questions (FAQ)

Is sec(2x)=(sec^2(x))/(1-tan^2(x)) ?
The answer to whether sec(2x)=(sec^2(x))/(1-tan^2(x)) is True