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2 years ago

Which expression is equivalent to sec^2x − 1? a. cot2x b. tan2x c. csc2x d. cos2x

2 answers:

4 0

Sec^2 x - 1 = tan^2x

Proof:
Sec^2x = 1+ tan^2x

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

<span>1/cos^2x - sin^2x/cos^2x = 1
</span>Using common denominator:
(1-sin^2x)/cos^2x = 1
sin^2x + cos^2 x = 1
cos^2 x = 1 - sin^2x
Substituting :
cos^2x/<span>cos^2x = 1
</span>1 = 1
Left hand side = right hand side

Artyom0805 [142]2 years ago

4 0

Answer:

option B tan²x

Step-by-step explanation:

We have to simplify the expression given as ( $sec^{2}$ x-1 )

Since sec x = $\frac{1}{cosx}$

so ( $\frac{1}{cosx}$ )² - 1 = $\frac{1}{cos^{2} x}$ - 1

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

= $\frac{sin^{2}x }{cos^{2}x}$

= tan²x

Therefore, option B tan²x is the answer.

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1 year ago

I GIVE BRAINLIEST PLS HELP

ioda

The answer is 8
Here's why:
${ ( \frac{( {6}^{7}) \times ( {3}^{3}) }{( {6}^{6}) \times ( {3}^{4} ) } )}^{3} = \\ ( \frac{6}{3} ) ^{3} = \\ \frac{216}{27} = 8$
The exponents are subtracted one from another when divided.
$\frac{ a ^{b} }{ {a}^{c} } = {a}^{b - c}$
We can look at the problem this way:
$( \frac{6^{7} }{6 ^{6} } \times \frac{3^{3} }{ {3}^{4} } ) = (6^{7 - 6} \times {3}^{3 - 4} ) = \\ ({6}^{1} \times {3}^{ - 1} )$
Since we have the power of -1 on the 3, we apply this rule:
${a}^{ - b} = \frac{1}{ {a}^{b} }$
Also this rule because we have the power of 1 on the 6:
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Then we get this:
$(6 \times \frac{1}{3} )^{3} = ( \frac{6}{3} )^{3}$
We apply the rule:
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We get this:
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3 years ago

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Answer:

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Step-by-step explanation:

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