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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

sec^{4} (α) - 1

Rewrite

sec^{4} (α) - 1

(sec^{2} (α))^{2} - 1^{2}

sec^{4} (α) - 1

Rewrite

1

1^{2}

= sec^{4} (α) - 1^{2}

Apply exponent rule:

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

sec^{4} (α) = (sec^{2} (α))^{2}

= (sec^{2} (α))^{2} - 1^{2}

= (sec^{2} (α))^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sec^{2} (α) + 1) (sec^{2} (α) - 1)

Factor

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

sec^{2} (α) - 1

Rewrite

1

1^{2}

= sec^{2} (α) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sec (α) + 1) (sec (α) - 1)

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

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

Factor

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

sec^{2} (α) - 1

Rewrite

1

1^{2}

= sec^{2} (α) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sec (α) + 1) (sec (α) - 1)

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

Cancel

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

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

Cancel the common factor:

sec (α) + 1

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

Cancel the common factor:

sec (α) - 1

= sec^{2} (α) + 1

= sec^{2} (α) + 1

= sec^{2} (α) + 1

= sec^{2} (α) + 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + sec ( x )}{csc ( x )} - tan (x) = sin (x)

2 sin (x) = \frac{4 cos ( x ) - 1}{tan ( x )}

cos (8 x) = cos^{2} (4 x) - sin^{2} (4 x)

tan (\frac{π}{2} - x) cot (x) = csc^{2} (x) - 1

cot^{2} (θ) cos (2 θ) = cot^{2} (θ) - 2 cos^{2} (θ)

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