Answer:
See Explanation
Step-by-step explanation:
Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified
Given
[tex]sec^4 x + sec^2 x tan^2 x - 2 tan^4 x[/tex]
Required
Simplify
[tex](sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2[/tex]
Represent [tex]sec^2x[/tex] with a
Represent [tex]tan^2x[/tex] with b
The expression becomes
[tex]a^2 + ab- 2 b^2[/tex]
Factorize
[tex]a^2 + 2ab -ab- 2 b^2[/tex]
[tex]a(a + 2b) -b(a+ 2 b)[/tex]
[tex](a -b) (a+ 2 b)[/tex]
Recall that
[tex]a = sec^2x[/tex]
[tex]b = tan^2x[/tex]
The expression [tex](a -b) (a+ 2 b)[/tex] becomes
[tex](sec^2x -tan^2x) (sec^2x+ 2 tan^2x)[/tex]
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In trigonometry
[tex]sec^2x =1 +tan^2x[/tex]
Subtract [tex]tan^2x[/tex] from both sides
[tex]sec^2x - tan^2x =1 +tan^2x - tan^2x[/tex]
[tex]sec^2x - tan^2x =1[/tex]
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Substitute 1 for [tex]sec^2x - tan^2x[/tex] in [tex](sec^2x -tan^2x) (sec^2x+ 2 tan^2x)[/tex]
[tex](1) (sec^2x+ 2 tan^2x)[/tex]
Open Bracket
[tex]sec^2x+ 2 tan^2x[/tex] ------------------This is an equivalence
[tex](secx)^2+ 2 (tanx)^2[/tex]
Solving further;
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In trigonometry
[tex]secx = \frac{1}{cosx}[/tex]
[tex]tanx = \frac{sinx}{cosx}[/tex]
Substitute the expressions for secx and tanx
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[tex](secx)^2+ 2 (tanx)^2[/tex] becomes
[tex](\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2[/tex]
Open bracket
[tex]\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})[/tex]
[tex]\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}[/tex]
Add Fraction
[tex]\frac{1 + 2sin^2x}{cos^2x}[/tex] ------------------------ This is another equivalence
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In trigonometry
[tex]sin^2x + cos^2x= 1[/tex]
Make [tex]sin^2x[/tex] the subject of formula
[tex]sin^2x= 1 - cos^2x[/tex]
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Substitute the expressions for [tex]1 - cos^2x[/tex] for [tex]sin^2x[/tex]
[tex]\frac{1 + 2(1 - cos^2x)}{cos^2x}[/tex]
Open bracket
[tex]\frac{1 + 2 - 2cos^2x}{cos^2x}[/tex]
[tex]\frac{3 - 2cos^2x}{cos^2x}[/tex] ---------------------- This is another equivalence
