Is tan^4(θ)-sec^4(θ)=1-2sec^2(θ) ?

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

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prove tan^4(θ)-sec^4(θ)=1-2sec^2(θ)

prove

tan^{4} (θ) - sec^{4} (θ) = 1 - 2 sec^{2} (θ)

True

Solution steps

tan^{4} (θ) - sec^{4} (θ) = 1 - 2 sec^{2} (θ)

Manipulating left side

tan^{4} (θ) - sec^{4} (θ)

Factor

- sec^{4} (θ) + tan^{4} (θ) : (tan^{2} (θ) + sec^{2} (θ)) (tan (θ) + sec (θ)) (tan (θ) - sec (θ))

- sec^{4} (θ) + tan^{4} (θ)

Rewrite

tan^{4} (θ) - sec^{4} (θ)

(tan^{2} (θ))^{2} - (sec^{2} (θ))^{2}

tan^{4} (θ) - sec^{4} (θ)

Apply exponent rule:

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

sec^{4} (θ) = (sec^{2} (θ))^{2}

= tan^{4} (θ) - (sec^{2} (θ))^{2}

Apply exponent rule:

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

tan^{4} (θ) = (tan^{2} (θ))^{2}

= (tan^{2} (θ))^{2} - (sec^{2} (θ))^{2}

= (tan^{2} (θ))^{2} - (sec^{2} (θ))^{2}

Apply Difference of Two Squares Formula:

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

(tan^{2} (θ))^{2} - (sec^{2} (θ))^{2} = (tan^{2} (θ) + sec^{2} (θ)) (tan^{2} (θ) - sec^{2} (θ))

= (tan^{2} (θ) + sec^{2} (θ)) (tan^{2} (θ) - sec^{2} (θ))

Factor

tan^{2} (θ) - sec^{2} (θ) : (tan (θ) + sec (θ)) (tan (θ) - sec (θ))

tan^{2} (θ) - sec^{2} (θ)

Apply Difference of Two Squares Formula:

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

tan^{2} (θ) - sec^{2} (θ) = (tan (θ) + sec (θ)) (tan (θ) - sec (θ))

= (tan (θ) + sec (θ)) (tan (θ) - sec (θ))

= (tan^{2} (θ) + sec^{2} (θ)) (tan (θ) + sec (θ)) (tan (θ) - sec (θ))

= (- sec (θ) + tan (θ)) (sec (θ) + tan (θ)) (sec^{2} (θ) + tan^{2} (θ))

Rewrite using trig identities

(- sec (θ) + tan (θ)) (sec (θ) + tan (θ)) (sec^{2} (θ) + tan^{2} (θ))

Use the Pythagorean identity:

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

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

= (- sec (θ) + tan (θ)) (sec (θ) + tan (θ)) (sec^{2} (θ) + sec^{2} (θ) - 1)

Add similar elements:

sec^{2} (θ) + sec^{2} (θ) = 2 sec^{2} (θ)

= (tan (θ) - sec (θ)) (sec (θ) + tan (θ)) (2 sec^{2} (θ) - 1)

Expand

(tan (θ) + sec (θ)) (tan (θ) - sec (θ)) : tan^{2} (θ) - sec^{2} (θ)

(tan (θ) + sec (θ)) (tan (θ) - sec (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = tan (θ), b = sec (θ)

= tan^{2} (θ) - sec^{2} (θ)

= (- 1 + 2 sec^{2} (θ)) (tan^{2} (θ) - sec^{2} (θ))

Use the Pythagorean identity:

sec^{2} (θ) = tan^{2} (θ) + 1

sec^{2} (θ) - tan^{2} (θ) = 1

= (- 1 + 2 sec^{2} (θ)) (- 1)

Simplify

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

(- 1 + 2 sec^{2} (θ)) (- 1)

Remove parentheses:

(- a) = - a

= - (- 1 + 2 sec^{2} (θ)) \cdot 1

Multiply:

(- 1 + 2 sec^{2} (θ)) \cdot 1 = (- 1 + 2 sec^{2} (θ))

= - (2 sec^{2} (θ) - 1)

Distribute parentheses

= - (- 1) - (2 sec^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= 1 - 2 sec^{2} (θ)

= 1 - 2 sec^{2} (θ)

= 1 - 2 sec^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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