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3 years ago

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At lunch time, Benjamin often borrows money from his friends to buy snacks in the school cafeteria. Benjamin borrowed $0.75 from

his friend Clyde five days last week to buy chips. Represent the amount Benjamin borrowed as the product of two rational numbers; then, determine how much Benjamin owed his friend last week.

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

<h2><em>3.75</em></h2>

Step-by-step explanation:

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Complete the identity.<br> 1) sec^4 x + sec^2 x tan^2 x - 2 tan^4 x = ?

Alecsey [184]

Answer:

See Explanation

Step-by-step explanation:

<em>Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified</em>

Given

$sec^4 x + sec^2 x tan^2 x - 2 tan^4 x$

Required

Simplify

$(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2$

Represent $sec^2x$ with a

Represent $tan^2x$ with b

The expression becomes

$a^2 + ab- 2 b^2$

Factorize

$a^2 + 2ab -ab- 2 b^2$

$a(a + 2b) -b(a+ 2 b)$

$(a -b) (a+ 2 b)$

Recall that

$a = sec^2x$

$b = tan^2x$

The expression $(a -b) (a+ 2 b)$ becomes

$(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

..............................................................................................................................

In trigonometry

$sec^2x =1 +tan^2x$

Subtract $tan^2x$ from both sides

$sec^2x - tan^2x =1 +tan^2x - tan^2x$

$sec^2x - tan^2x =1$

..............................................................................................................................

Substitute 1 for $sec^2x - tan^2x$ in $(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)$

$(1) (sec^2x+ 2 tan^2x)$

Open Bracket

$sec^2x+ 2 tan^2x$ ------------------This is an equivalence

$(secx)^2+ 2 (tanx)^2$

Solving further;

................................................................................................................................

In trigonometry

$secx = \frac{1}{cosx}$

$tanx = \frac{sinx}{cosx}$

Substitute the expressions for secx and tanx

................................................................................................................................

$(secx)^2+ 2 (tanx)^2$ becomes

$(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2$

Open bracket

$\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})$

$\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}$

Add Fraction

$\frac{1 + 2sin^2x}{cos^2x}$ ------------------------ This is another equivalence

................................................................................................................................

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$sin^2x + cos^2x= 1$

Make $sin^2x$ the subject of formula

$sin^2x= 1 - cos^2x$

................................................................................................................................

Substitute the expressions for $1 - cos^2x$ for $sin^2x$

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

Open bracket

$\frac{1 + 2 - 2cos^2x}{cos^2x}$

$\frac{3 - 2cos^2x}{cos^2x}$ ---------------------- This is another equivalence

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Step-by-step explanation:

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Step-by-step explanation:

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