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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

\frac{1 - tan ^{4} ( θ )}{tan ^{2} ( θ ) + 1} : - (tan (θ) + 1) (tan (θ) - 1)

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

Factor

1 - tan^{4} (θ) : - (tan^{2} (θ) + 1) (tan (θ) + 1) (tan (θ) - 1)

1 - tan^{4} (θ)

Factor out common term

- 1

= - (tan^{4} (θ) - 1)

Factor

tan^{4} (θ) - 1 : (tan^{2} (θ) + 1) (tan (θ) + 1) (tan (θ) - 1)

tan^{4} (θ) - 1

Rewrite

tan^{4} (θ) - 1

(tan^{2} (θ))^{2} - 1^{2}

tan^{4} (θ) - 1

Rewrite

1

1^{2}

= tan^{4} (θ) - 1^{2}

Apply exponent rule:

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

tan^{4} (θ) = (tan^{2} (θ))^{2}

= (tan^{2} (θ))^{2} - 1^{2}

= (tan^{2} (θ))^{2} - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (tan^{2} (θ) + 1) (tan^{2} (θ) - 1)

Factor

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

tan^{2} (θ) - 1

Rewrite

1

1^{2}

= tan^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (tan (θ) + 1) (tan (θ) - 1)

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

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

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

Cancel the common factor:

tan^{2} (θ) + 1

= - (tan (θ) + 1) (tan (θ) - 1)

= - (tan (θ) + 1) (tan (θ) - 1)

= - (tan (θ) + 1) (tan (θ) - 1)

Expand

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

- (- 1 + tan (θ)) (1 + tan (θ))

Expand

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

(- 1 + tan (θ)) (1 + tan (θ))

Apply Difference of Two Squares Formula:

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

a = tan (θ), b = 1

= tan^{2} (θ) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= tan^{2} (θ) - 1

= - (tan^{2} (θ) - 1)

Distribute parentheses

= - (tan^{2} (θ)) - (- 1)

Apply minus-plus rules

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

= - tan^{2} (θ) + 1

= 1 - tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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