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  1. #1
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    Default Quant Thread for CAT 2009 (Algebra)

    Question : If 2sinA.cosB.sinC = sinB.sin(A+C),then tanA,tanB,tanC are in ...

    Options :

    1)A.P.
    2)G.P.
    3)H.P.
    4)A.G.P.
    5)No relation

  2. #2
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    Quote Originally Posted by baburaoapate View Post
    Question : If 2sinA.cosB.sinC = sinB.sin(A+C),then tanA,tanB,tanC are in ...

    Options :

    1)A.P.
    2)G.P.
    3)H.P.
    4)A.G.P.
    5)No relation
    Is this question right becuase you need some restriction on value of A,B and C????
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    Cool

    The condition itself restricts the value of A, B and C

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    Quote Originally Posted by baburaoapate View Post
    Question : If 2sinA.cosB.sinC = sinB.sin(A+C),then tanA,tanB,tanC are in ...

    Options :

    1)A.P.
    2)G.P.
    3)H.P.
    4)A.G.P.
    5)No relation
    let i take A=B=C =0 then the condition is met and then we have tanA = tanB=tanC = 0 ...which can be AP,GP or whatever...

    any other takers for this question..

    @baburao..do u have the sol for the same
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    Regarding that question from trigonometry, by expanding the terms I've come across 3 cases all of which leads to A=B=C=0(considering the 1st qudrant values only).

    It'll be much appreciated if baburao can explain the method and solution to the problem

  6. #6
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    Default Quant Thread for CAT 2009 (Algebra)

    post all your queries , questions regarding equations, function , sequences here
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    Question

    Q: Log[(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = ?

    a) 0 b) 2 c) 1 d) infinity
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    Quote Originally Posted by sumit2goody View Post
    Q: Log[(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = ?

    a) 0 b) 2 c) 1 d) infinity

    Ans is a)0

  9. #9
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    Quote Originally Posted by sumit2goody View Post
    Q: Log[(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = ?

    a) 0 b) 2 c) 1 d) infinity
    it's a) 0

    tanAtanB = 1-(tanA+tan B)/tan(A+B)
    Putting A = 2 and B= 88

    tan2tan88 = 1-(tan2+tan88)/tan 90 = 1-0 =1(tan 90 = infinity)
    we get 1 for all the pairs and we are left with tan 45 with no pair.

    Now tan 45 = 1

    so it comes down to log(1.1.1...) = log 1 = 0.

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    Quote Originally Posted by first_timer View Post
    it's a) 0

    tanAtanB = 1-(tanA+tan B)/tan(A+B)
    Putting A = 2 and B= 88

    tan2tan88 = 1-(tan2+tan88)/tan 90 = 1-0 =1(tan 90 = infinity)
    we get 1 for all the pairs and we are left with tan 45 with no pair.

    Now tan 45 = 1

    so it comes down to log(1.1.1...) = log 1 = 0.
    Right nay other methods ???
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    Quote Originally Posted by sumit2goody View Post
    Q: Log[(tan 2)(tan 4)(tan 6)....(tan 86)(tan 88)] = ?

    a) 0 b) 2 c) 1 d) infinity
    it's option a..
    used the same method as of first_timer...

  12. #12
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    Oh my God! Looking at trignometery gives me heartattack! Is school level maths enogh to solve such sums? So that I cant start revising it.

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    Cant figure out the answer!!! And to think of it trigonometry was my favourite topic in school Is the difficulty level oo high for this question?

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    Quote Originally Posted by JNeetu View Post
    Oh my God! Looking at trignometery gives me heartattack! Is school level maths enogh to solve such sums? So that I cant start revising it.
    If trigonometry is such a big issue for u i think u should revise ur school level trigo and should clear ur basic concepts as soon as possible.
    after this, u r more than half done.

    if u r referring any book for ur quant prep, there is no need to go through school books; u can study from that book only. All the prep books include basic concepts plus short-cuts to save ur precious time during the exam

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    Answer is option 1(0)

    i found this using foll method.

    tanx=cot(90-x)
    tan2=cot(90-2)=cot88=1/(tan88)
    hence tan2.tan88=1
    similarly tan4.tan86=1
    thus log(tan2.tan4....tan86.tan88)=log(1)=0

  16. The Following User Says Thank You to raju89 For This Useful Post:

    BelltheC@T (04-Apr-09)

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    Quote Originally Posted by SimiNaik View Post
    Cant figure out the answer!!! And to think of it trigonometry was my favourite topic in school Is the difficulty level oo high for this question?
    Hi simi,

    see this question uses very basic concept of trigo...

    one is that tan x = cot(90-x)

    and you can also solve it using tan(A+B) = tan A + tanB / 1 - tanAtanB

    so see both are very basic concepts ..just revise them and try practicing questions you will crack your fav topic
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  18. The Following User Says Thank You to sumit2goody For This Useful Post:

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    Quote Originally Posted by sumit2goody View Post
    Right nay other methods ???

    Tan 88 = Cot 2 [Tan A = Cot (90 - A)]

    And Tan 2 . Cot 2 = 1

    So, we form pairs like :
    Tan 4 . Tan 86 = 1
    Tan 6 . Tan 84 = 1
    Tan 8 . Tan 82 = 1...

    And Log 1 = 0

  20. #18
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    Cudnt do thuis sum. So I saw the method used by you all. Now I got it. Will practic more sums now.

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    Smile thank you nipunkesari

    Thank you nipunkesari. I will practice from school books and will also practice from other CAT level books. Will clear my basics first so that I can do the sums properly.

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    Talking THANK YOU Sumit!!!

    Thanks Sumit. Yeah that really was easy. Those formulae I learnt in class 10th. Maybe I just dint see it. Now there are hopes of trigo remaining my fav. Topic

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