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Lesechka [4]
3 years ago
14

TRIGONOMETRY , can you solve this plz:(even if you know one of them,plz help:)​

Mathematics
1 answer:
daser333 [38]3 years ago
5 0

Answer:

See below

Step-by-step explanation:

\frac{1}{1-cos A} - \frac{1}{1+cos A} =2 cosec A cot A \\  \\ L. H. S.= \frac{1}{1-cos A} - \frac{1}{1+cos A} \\  \\ = \frac{1+cos A - (1 - cos A)}{(1-cos A)(1+cos A)}\\  \\ = \frac{1+cos A - 1  +  cos A}{1-cos^{2}  A}\\  \\ = \frac{2cos A }{sin^{2}  A}\\  \\ =2 .\frac{1}{sin A}.\frac{cos A }{sin  A}.\\  \\  = 2.cosec \: A.cot \: A \\  \\  = R.H.S

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Find the number of possible ways one can choose marbles from a bag containing three marbles: a) if the order is important b) if
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Answer:

There are 3 marbles, that we assume are different.

Let's define them as marble 1, marble 2, and marble 3.

a) If the order is important, and we select the 3 marbles, then:

for the first marble, we have 3 options.

For the second marble, we have 2 options (because one is already taken)

for the third marble is only one option.

The total number of combinations is equal to the product between the numbers of options, so in this case, we have:

C = 3*2*1 = 6 combinations.

Now, if we only select 2 marbles from the bag, we have:

for the first marble, we have 3 options.

For the second marble, we have 2 options

Here the number of different combinations is:

C = 3*2 = 6 combinations.

And if we select only one marble from the bag, we have:

for the first marble, we have 3 options.

Then here are only 3 combinations.

the total number of combinations is:

C' = 6 + 6 + 3 = 15 different combinations.

b) If the order is not important, then there is only one combination when we draw the 3 marbles.

When we draw one marble there are 3 combinations (one for each marble)

When we draw two marbles, again there are 3 combinations (one for each marble that we do not draw)

then the total number of combinations in this case is:

C = 1 + 3 + 3 = 7

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Step-by-step explanation:

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