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    Simplify :

    tg x + cotg x 􀀀- 1/(sin x cos x)
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    (Original post by priyaroy789)
    Simplify :

    tg x + cotg x 􀀀- 1/(sin x cos x)
    What is that meant to say? tan x + cot x = -sec x cosec x?
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    (Original post by Zacken)
    What is that meant to say? tan x + cot x = -sec x cosec x?
    Thats exactly what the question says but i just looked it up and found that tg x is the same as tan x
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    (Original post by priyaroy789)
    Thats exactly what the question says but i just looked it up and found that tg x is the same as tan x
    Write tangent and cotangent in their sin/cos forms and then common denominator to victory.
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    (Original post by Zacken)
    Write tangent and cotangent in their sin/cos forms and then common denominator to victory.
    Cheers, do you think you'll be able to help me out with another question while you're online?
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    (Original post by priyaroy789)
    Cheers, do you think you'll be able to help me out with another question while you're online?
    How many dierent nine-digit numbers can be created by moving digits in the number 123454321?
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    (Original post by priyaroy789)
    How many dierent nine-digit numbers can be created by moving digits in the number 123454321?
    This is a combination problem.
    See how many different combinations you can make and then see how many repeating digits you have!
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    (Original post by dan94adibi)
    This is a combination problem.
    See how many different combinations you can make and then see how many repeating digits you have!
    yup, is there a method/formula to figure it out?
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    (Original post by priyaroy789)
    yup, is there a method/formula to figure it out?
    Yes. I'll give you a hint.
    In 123454321 you have two 1s, two 2s, two 3s, and two 4s. Thus, some of the combinations will yield the same answer and you will have to divide them from the total combinations.
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    (Original post by dan94adibi)
    Yes. I'll give you a hint.
    In 123454321 you have two 1s, two 2s, two 3s, and two 4s. Thus, some of the combinations will yield the same answer and you will have to divide them from the total combinations.
    so like 9!/(2!*2!*2!*2!), but what about the five?
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    (Original post by priyaroy789)
    so like 9!/(2!*2!*2!*2!), but what about the five?
    That would be an extra division by 1! which is just 1 so not necessary.
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    The five would appear as this:

    \displaystyle \dfrac{9!}{2!*2!*2!*2!*1!}

    but since 1! = 1 it's not necessary to consider it. There's only one 5, so there will be no repeats because of it.
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    (Original post by priyaroy789)
    so like 9!/(2!*2!*2!*2!), but what about the five?
    Yup the 5 only occurs once. Essentially, what you are doing is crossing out those that are 'double counted'.
    For example:
    1(a)1(b)232454
    is the same as
    1(b)1(a)232454

    Its just that the two ones have swapped position but really they're both the same number. So, the total combination is 22,680.
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    (Original post by priyaroy789)
    Simplify :

    tg x + cotg x 􀀀- 1/(sin x cos x)
    Thank You Guys For Replying
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    (Original post by priyaroy789)
    Thank You Guys For Replying
    No worries.
 
 
 
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