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2 years ago

Plz help me well mark brainliest if you are correct!

Mathematics

2 answers:

ludmilkaskok [199]2 years ago

8 0

Answer:

-4

Step-by-step explanation:

First combine 2/3 - 4, the quantity enclosed by absolute value symbols | :

The LCD heree is 3. Thus, 2/3 - 4 can be rewritten as 2/3 - 12/3, or -10/3.

Next, take the absolute value of this -10/3: we get 10/3.

Then we have -10/3 divided by 5/6:

-10

-------

--------------

----

Here we're dividing by a fraction. An equivalent expression involves inverting 5/6 and multiplying the result by -10/3:

-10 6

----- * -----

3 5

This reduces to -2(2), or -4

Luba_88 [7]2 years ago

4 0

The correct answer is -4 I am 100 percent sure of it

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Prove the following

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Answer:

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

<h2 /><h2><u>Consider</u></h2>

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)$

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx$

$\rm \: {cos \: (2\pi + x) }$

$\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx$

$\rm \: cot(2\pi + x) \: = \: cotx$

So, on substituting all these values, we get

$\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}$

$\rm \: = \: 1=1$

<h2>Hence,</h2>

$\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

sin (90°-θ) = cos θ
cos (90°-θ) = sin θ
tan (90°-θ) = cot θ
csc (90°-θ) = sec θ
sec (90°-θ) = csc θ
cot (90°-θ) = tan θ
sin (90°+θ) = cos θ
cos (90°+θ) = -sin θ
tan (90°+θ) = -cot θ
csc (90°+θ) = sec θ
sec (90°+θ) = -csc θ
cot (90°+θ) = -tan θ
sin (180°-θ) = sin θ
cos (180°-θ) = -cos θ
tan (180°-θ) = -tan θ
csc (180°-θ) = csc θ
sec (180°-θ) = -sec θ
cot (180°-θ) = -cot θ
sin (180°+θ) = -sin θ
cos (180°+θ) = -cos θ
tan (180°+θ) = tan θ
csc (180°+θ) = -csc θ
sec (180°+θ) = -sec θ
cot (180°+θ) = cot θ
sin (270°-θ) = -cos θ
cos (270°-θ) = -sin θ
tan (270°-θ) = cot θ
csc (270°-θ) = -sec θ
sec (270°-θ) = -csc θ
cot (270°-θ) = tan θ
sin (270°+θ) = -cos θ
cos (270°+θ) = sin θ
tan (270°+θ) = -cot θ
csc (270°+θ) = -sec θ
sec (270°+θ) = cos θ
cot (270°+θ) = -tan θ

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3 years ago

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4 years ago

I need help please.

makkiz [27]

Answer:

see explanation

Step-by-step explanation:

Using the rules of radicals/ exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$

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3 years ago

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