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Popular Trigonometry >

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

  • Pre Algebra
  • Algebra
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Solution

prove

Solution

True
Solution steps
tan4(x)−sec4(x)=−2sec2(x)+1
Manipulating left sidetan4(x)−sec4(x)
Factor −sec4(x)+tan4(x):(tan2(x)+sec2(x))(tan(x)+sec(x))(tan(x)−sec(x))
−sec4(x)+tan4(x)
Rewrite tan4(x)−sec4(x) as (tan2(x))2−(sec2(x))2
tan4(x)−sec4(x)
Apply exponent rule: abc=(ab)csec4(x)=(sec2(x))2=tan4(x)−(sec2(x))2
Apply exponent rule: abc=(ab)ctan4(x)=(tan2(x))2=(tan2(x))2−(sec2(x))2
=(tan2(x))2−(sec2(x))2
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(tan2(x))2−(sec2(x))2=(tan2(x)+sec2(x))(tan2(x)−sec2(x))=(tan2(x)+sec2(x))(tan2(x)−sec2(x))
Factor tan2(x)−sec2(x):(tan(x)+sec(x))(tan(x)−sec(x))
tan2(x)−sec2(x)
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)tan2(x)−sec2(x)=(tan(x)+sec(x))(tan(x)−sec(x))=(tan(x)+sec(x))(tan(x)−sec(x))
=(tan2(x)+sec2(x))(tan(x)+sec(x))(tan(x)−sec(x))
=(−sec(x)+tan(x))(sec(x)+tan(x))(sec2(x)+tan2(x))
Rewrite using trig identities
(−sec(x)+tan(x))(sec(x)+tan(x))(sec2(x)+tan2(x))
Use the Pythagorean identity: tan2(x)+1=sec2(x)tan2(x)=sec2(x)−1=(−sec(x)+tan(x))(sec(x)+tan(x))(sec2(x)+sec2(x)−1)
Add similar elements: sec2(x)+sec2(x)=2sec2(x)=(tan(x)−sec(x))(sec(x)+tan(x))(2sec2(x)−1)
Expand (tan(x)+sec(x))(tan(x)−sec(x)):tan2(x)−sec2(x)
(tan(x)+sec(x))(tan(x)−sec(x))
Apply Difference of Two Squares Formula: (a+b)(a−b)=a2−b2a=tan(x),b=sec(x)=tan2(x)−sec2(x)
=(−1+2sec2(x))(tan2(x)−sec2(x))
Use the Pythagorean identity: sec2(x)=tan2(x)+1sec2(x)−tan2(x)=1=(−1+2sec2(x))(−1)
Simplify (−1+2sec2(x))(−1):1−2sec2(x)
(−1+2sec2(x))(−1)
Remove parentheses: (−a)=−a=−(−1+2sec2(x))⋅1
Multiply: (−1+2sec2(x))⋅1=(−1+2sec2(x))=−(2sec2(x)−1)
Distribute parentheses=−(−1)−(2sec2(x))
Apply minus-plus rules−(−a)=a,−(a)=−a=1−2sec2(x)
=1−2sec2(x)
=1−2sec2(x)
=−2sec2(x)+1
We showed that the two sides could take the same form⇒True

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Frequently Asked Questions (FAQ)

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

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