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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

sec^{4} (x) - 1

Rewrite

sec^{4} (x) - 1

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

sec^{4} (x) - 1

Rewrite

1

1^{2}

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

sec^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (sec (x) + 1) (sec (x) - 1)

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

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

Factor

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

sec^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (sec (x) + 1) (sec (x) - 1)

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

Cancel

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

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

Cancel the common factor:

sec (x) + 1

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

Cancel the common factor:

sec (x) - 1

= sec^{2} (x) + 1

= sec^{2} (x) + 1

= sec^{2} (x) + 1

Use the Pythagorean identity:

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

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

Simplify

= tan^{2} (x) + 2

= tan^{2} (x) + 2

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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