Is (1-tan^4(x))/(1-tan^2(x))=sec^2(x) ?

The answer to whether (1-tan^4(x))/(1-tan^2(x))=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 (1-tan^4(x))/(1-tan^2(x))=sec^2(x)

Solution

prove

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

Solution

True

Solution steps

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

Manipulating left side

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )}

Simplify

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} : tan^{2} (x) + 1

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )}

Factor

1 - tan^{4} (x) : - (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

1 - tan^{4} (x)

Factor out common term

- 1

= - (tan^{4} (x) - 1)

Factor

tan^{4} (x) - 1 : (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

tan^{4} (x) - 1

Rewrite

tan^{4} (x) - 1

(tan^{2} (x))^{2} - 1^{2}

tan^{4} (x) - 1

Rewrite

1

1^{2}

= tan^{4} (x) - 1^{2}

Apply exponent rule:

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

tan^{4} (x) = (tan^{2} (x))^{2}

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(tan^{2} (x))^{2} - 1^{2} = (tan^{2} (x) + 1) (tan^{2} (x) - 1)

= (tan^{2} (x) + 1) (tan^{2} (x) - 1)

Factor

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - 1^{2} = (tan (x) + 1) (tan (x) - 1)

= (tan (x) + 1) (tan (x) - 1)

= (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

= - (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

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

Factor

1 - tan^{2} (x) : - (tan (x) + 1) (tan (x) - 1)

1 - tan^{2} (x)

Factor out common term

- 1

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

Factor

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - 1^{2} = (tan (x) + 1) (tan (x) - 1)

= (tan (x) + 1) (tan (x) - 1)

= - (tan (x) + 1) (tan (x) - 1)

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

Cancel

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

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

Cancel the common factor:

tan (x) + 1

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

Cancel the common factor:

tan (x) - 1

= tan^{2} (x) + 1

= tan^{2} (x) + 1

= 1 + tan^{2} (x)

Rewrite using trig identities

1 + tan^{2} (x)

Use the Pythagorean identity:

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

prove arcsin(x)=csc(x)

prove cos^4(A)-sin^4(A)=cos(2A)

prove (1+sec^2(x))/(sec^2(x))=1+cos^2(x)

prove tan(θ)+1=sec(θ)

Frequently Asked Questions (FAQ)

Is (1-tan^4(x))/(1-tan^2(x))=sec^2(x) ?
The answer to whether (1-tan^4(x))/(1-tan^2(x))=sec^2(x) is True