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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

\frac{csc ^{4} ( x ) - 1}{csc ^{2} ( x ) - 1} : csc^{2} (x) + 1

\frac{csc ^{4} ( x ) - 1}{csc ^{2} ( x ) - 1}

Factor

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

csc^{4} (x) - 1

Rewrite

csc^{4} (x) - 1

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

csc^{4} (x) - 1

Rewrite

1

1^{2}

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

Apply exponent rule:

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

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

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

csc^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (csc (x) + 1) (csc (x) - 1)

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

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

Factor

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

csc^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (csc (x) + 1) (csc (x) - 1)

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

Cancel

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

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

Cancel the common factor:

csc (x) + 1

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

Cancel the common factor:

csc (x) - 1

= csc^{2} (x) + 1

= csc^{2} (x) + 1

= csc^{2} (x) + 1

Use the Pythagorean identity:

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

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

Simplify

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

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

Group like terms

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

Add the numbers:

1 + 1 = 2

= cot^{2} (x) + 2

= cot^{2} (x) + 2

= cot^{2} (x) + 2

= 2 + cot^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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