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Oksanka [162]
3 years ago
6

Let tan(x)=25tan(x)=25 . What is the value of tan(π−x)tan(π−x) ?

Mathematics
2 answers:
galina1969 [7]3 years ago
7 0
We are given that tanx = 25.

We are also given tan(π−x).

Apply distributive rule to
tan(π−x).

tanπ − tanx

Take it from here knowing that tanx has been given to be 25.

Ghella [55]3 years ago
4 0
Given tan(x)=25
we need to find tan(π-x)=cot(x) = 1/tan(x) = 1/25

Answer: tan(π-x)=1/25.
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wariber [46]

Answer: 8 days

Step-by-step explanation:

6 miles has been paved already.

Number of miles left = 30 - 6

= 24 miles

If they are paving at 3 miles a day, the days they will take to finish is:

= 24/3

= 8 days

5 0
2 years ago
Read 2 more answers
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

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3 years ago
I need help with 1-4
worty [1.4K]
1) the longest side is 5cm and the perimeter is 12cm
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Alenkinab [10]

Answer:

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2 years ago
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amm1812

Answer:

5

Step-by-step explanation:

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√25+0

=5

5 0
2 years ago
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