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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

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

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

Rewrite

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

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

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

= (sec (x) + tan (x)) (sec (x) - tan (x))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

Expand

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

(sec (x) + tan (x)) (sec (x) - tan (x))

Apply Difference of Two Squares Formula:

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

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

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

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (- x) = - sin (x)

p ro v e sin (x + π) = - sin (x)

p ro v e \frac{cos ( 2 X )}{sin ( X )} + \frac{sin ( 2 X )}{cos ( X )} = csc (X)

p ro v e tan^{5} (x) = tan^{3} (x) sec^{2} (x) - tan^{3} (x)

p ro v e \frac{cos ( θ )}{1 - sin ( θ )} = sec (θ) + tan (θ)

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