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

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

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prove tan^4(k)-sec^4(k)=1-2sec^2(k)

prove

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

True

Solution steps

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

Manipulating left side

tan^{4} (k) - sec^{4} (k)

Factor

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

- sec^{4} (k) + tan^{4} (k)

Rewrite

tan^{4} (k) - sec^{4} (k)

(tan^{2} (k))^{2} - (sec^{2} (k))^{2}

tan^{4} (k) - sec^{4} (k)

Apply exponent rule:

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

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

= tan^{4} (k) - (sec^{2} (k))^{2}

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

tan^{2} (k) - sec^{2} (k)

Apply Difference of Two Squares Formula:

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

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

= (tan (k) + sec (k)) (tan (k) - sec (k))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Add similar elements:

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

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

Expand

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

(tan (k) + sec (k)) (tan (k) - sec (k))

Apply Difference of Two Squares Formula:

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

a = tan (k), b = sec (k)

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

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

Use the Pythagorean identity:

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

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

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

Simplify

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

(- 1 + 2 sec^{2} (k)) (- 1)

Remove parentheses:

(- a) = - a

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

Multiply:

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e 1 - tan^{4} (x) = 2 sec^{2} (x) - sec^{4} (x)

p ro v e sin (π) = sin (- π)

p ro v e csc (t) cot (t) = \frac{1 + cot ^{2} ( t )}{sec ( t )}

p ro v e csc (a) sec (a) = 2 csc (2 a)

p ro v e tan^{2} (2 x) + 1 = sec^{2} (2 x)

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