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Popular Trigonometry >

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

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Solution

prove csc4(x)−cot4(x)=2cot2(x)+1

Solution

True
Solution steps
csc4(x)−cot4(x)=2cot2(x)+1
Manipulating left sidecsc4(x)−cot4(x)
Factor −cot4(x)+csc4(x):(csc2(x)+cot2(x))(csc(x)+cot(x))(csc(x)−cot(x))
−cot4(x)+csc4(x)
Rewrite csc4(x)−cot4(x) as (csc2(x))2−(cot2(x))2
csc4(x)−cot4(x)
Apply exponent rule: abc=(ab)ccot4(x)=(cot2(x))2=csc4(x)−(cot2(x))2
Apply exponent rule: abc=(ab)ccsc4(x)=(csc2(x))2=(csc2(x))2−(cot2(x))2
=(csc2(x))2−(cot2(x))2
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(csc2(x))2−(cot2(x))2=(csc2(x)+cot2(x))(csc2(x)−cot2(x))=(csc2(x)+cot2(x))(csc2(x)−cot2(x))
Factor csc2(x)−cot2(x):(csc(x)+cot(x))(csc(x)−cot(x))
csc2(x)−cot2(x)
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)csc2(x)−cot2(x)=(csc(x)+cot(x))(csc(x)−cot(x))=(csc(x)+cot(x))(csc(x)−cot(x))
=(csc2(x)+cot2(x))(csc(x)+cot(x))(csc(x)−cot(x))
=(−cot(x)+csc(x))(cot(x)+csc(x))(cot2(x)+csc2(x))
Rewrite using trig identities
(−cot(x)+csc(x))(cot(x)+csc(x))(cot2(x)+csc2(x))
Expand (csc(x)+cot(x))(csc(x)−cot(x)):csc2(x)−cot2(x)
(csc(x)+cot(x))(csc(x)−cot(x))
Apply Difference of Two Squares Formula: (a+b)(a−b)=a2−b2a=csc(x),b=cot(x)=csc2(x)−cot2(x)
=(csc2(x)−cot2(x))(cot2(x)+csc2(x))
Use the Pythagorean identity: csc2(x)=1+cot2(x)csc2(x)−cot2(x)=1=1⋅(cot2(x)+csc2(x))
Simplify=cot2(x)+csc2(x)
Use the Pythagorean identity: csc2(x)=1+cot2(x)=cot2(x)+1+cot2(x)
Simplify cot2(x)+1+cot2(x):2cot2(x)+1
cot2(x)+1+cot2(x)
Group like terms=cot2(x)+cot2(x)+1
Add similar elements: cot2(x)+cot2(x)=2cot2(x)=2cot2(x)+1
=2cot2(x)+1
=2cot2(x)+1
We showed that the two sides could take the same form⇒True

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Frequently Asked Questions (FAQ)

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

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