Is csc^4(x)-cot^4(x)=2csc^2(x)-1 ?

The answer to whether csc^4(x)-cot^4(x)=2csc^2(x)-1 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove csc^4(x)-cot^4(x)=2csc^2(x)-1

prove

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) - 1

True

Solution steps

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) - 1

Manipulating left side

csc^{4} (x) - cot^{4} (x)

Factor

- cot^{4} (x) + csc^{4} (x) : (csc^{2} (x) + cot^{2} (x)) (csc (x) + cot (x)) (csc (x) - cot (x))

- cot^{4} (x) + csc^{4} (x)

Rewrite

csc^{4} (x) - cot^{4} (x)

(csc^{2} (x))^{2} - (cot^{2} (x))^{2}

csc^{4} (x) - cot^{4} (x)

Apply exponent rule:

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

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

= csc^{4} (x) - (cot^{2} (x))^{2}

Apply exponent rule:

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

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

= (csc^{2} (x))^{2} - (cot^{2} (x))^{2}

= (csc^{2} (x))^{2} - (cot^{2} (x))^{2}

Apply Difference of Two Squares Formula:

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

(csc^{2} (x))^{2} - (cot^{2} (x))^{2} = (csc^{2} (x) + cot^{2} (x)) (csc^{2} (x) - cot^{2} (x))

= (csc^{2} (x) + cot^{2} (x)) (csc^{2} (x) - cot^{2} (x))

Factor

csc^{2} (x) - cot^{2} (x) : (csc (x) + cot (x)) (csc (x) - cot (x))

csc^{2} (x) - cot^{2} (x)

Apply Difference of Two Squares Formula:

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

csc^{2} (x) - cot^{2} (x) = (csc (x) + cot (x)) (csc (x) - cot (x))

= (csc (x) + cot (x)) (csc (x) - cot (x))

= (csc^{2} (x) + cot^{2} (x)) (csc (x) + cot (x)) (csc (x) - cot (x))

= (- cot (x) + csc (x)) (cot (x) + csc (x)) (cot^{2} (x) + csc^{2} (x))

Rewrite using trig identities

(- cot (x) + csc (x)) (cot (x) + csc (x)) (cot^{2} (x) + csc^{2} (x))

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= (- cot (x) + csc (x)) (cot (x) + csc (x)) (csc^{2} (x) - 1 + csc^{2} (x))

Simplify

csc^{2} (x) - 1 + csc^{2} (x) : 2 csc^{2} (x) - 1

csc^{2} (x) - 1 + csc^{2} (x)

Group like terms

= csc^{2} (x) + csc^{2} (x) - 1

Add similar elements:

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

= 2 csc^{2} (x) - 1

= (csc (x) - cot (x)) (cot (x) + csc (x)) (2 csc^{2} (x) - 1)

Expand

(csc (x) + cot (x)) (csc (x) - cot (x)) : csc^{2} (x) - cot^{2} (x)

(csc (x) + cot (x)) (csc (x) - cot (x))

Apply Difference of Two Squares Formula:

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

a = csc (x), b = cot (x)

= csc^{2} (x) - cot^{2} (x)

= (- 1 + 2 csc^{2} (x)) (csc^{2} (x) - cot^{2} (x))

Use the Pythagorean identity:

csc^{2} (x) = 1 + cot^{2} (x)

csc^{2} (x) - cot^{2} (x) = 1

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

Simplify

= - 1 + 2 csc^{2} (x)

= - 1 + 2 csc^{2} (x)

= 2 csc^{2} (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

csc^{4} (x) - 2 csc^{2} (x) + 1 = cot^{4} (x)

\frac{cos ( x )}{1 - sin ( x )} + \frac{1 - sin ( x )}{cos ( x )} = 2 sec (x)

\frac{tan ( θ ) cot ( θ )}{csc ( θ )} = sin (θ)

(1 - cos (x)) (1 + cos (x)) = \frac{1}{csc ^{2} ( x )}

csc (x) cos (x) = cot (x)

Is csc^4(x)-cot^4(x)=2csc^2(x)-1 ?
The answer to whether csc^4(x)-cot^4(x)=2csc^2(x)-1 is True