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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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)

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)

= \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)

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)

= \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

csc (β) - csc (β) cos (2 β) = 2 \cdot sin (β)

cos (27 0^{\circ} - θ) = - sin (θ)

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

cot (2 a) = \frac{1}{2} (cot (a) - tan (a))

sec (t) = \frac{cot ( t ) + tan ( t )}{csc ( t )}

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