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mel-nik [20]
3 years ago
7

Prove the following

Mathematics
2 answers:
fomenos3 years ago
7 0

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)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\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
salantis [7]3 years ago
4 0

Step-by-step explanation:

Hence,

\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}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A
alexgriva [62]

Answer: A) Option B

B) 22.3 minutes.

Step-by-step explanation:

The temperature decreases in an exponential decrease, this means that we can write the equation as:

T(x) = A*r^x + B

Where A is the difference between the initial temperature and the room temperature, B is the room temperature, and r is a positive number smaller than 1, that says "how fast" the temperature decreases (and x is the variable, in units of time)

We know that the initial temperature

A = 200° - 72° = 128°

B = 72°

T(x) = 128°r^x + 72°

The only option with those two values is option B.

T(x) =  128(0.989)^x + 72

We should check that when x = 2m, the temperature must be 197°

T(2m) = 128*(0.989)^2 + 72 = 197.2° (that we can round down to 197°)

So this equation is correct

Now we want to find the time such the temperature is 172°F

then:

172° = 128°(0.989)^x + 72°

100° = 128°(0.989)^x

100/128 = (0.989)^x

x = ln(100/128)/ln(0.989) = 22.3 minutes.

Let's check in our equation:

T(22.3) =  128°(0.989)^22.3 + 72 = 172°

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2x + 46 = 7x + 6

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