In which of the following intervals does the trigonometric inequality sec(x) < cot(x) always hold true.

Mathematics

1 answer:

Ray Of Light [21]1 year ago

8 0

The inequality for sec (x) < cot (x) is; π/2 < x < π

<h3>How to express trigonometric inequality?</h3>

We are given that;

We want to find the intervals that the trigonometric inequality sec (x) < cot (x) always hold true.

This can also be expressed as;

1/cos (x) < 1/tan (x)

Now, this can happen only in the quadrant where tan (x) is negative and cos x is positive which is in fourth quadrant where;

π/2 < x < π

The inequality for sec (x) < cot (x) is; π/2 < x < π

Read more about Trigonometric Inequality at; brainly.com/question/12094532

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Solve for x. <br> 25x=65<br> x = 1225 <br> x = 125<br> x = 3 <br> x = 6

masya89 [10]

25x=65

We divide 25 to cancel it out.

65/25=2.6

I would assume 3, because 2.6 is 3 when rounded up.

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Gelneren [198K]

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Hello,

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Step-by-step explanation:

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