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Math Help - Worked Solution does not explain why tan x = - √15 is rejected

  1. #1
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    Worked Solution does not explain why tan x = - √15 is rejected

    Can any kind soul help me in the below?

    1. If x is obtuse and 2 tan^2 x = 5 sec x +10, find the value of tan x without using calculator.

    Answer:

    obtuse : 90deg < x< 180 deg
    2 tan^2 x = 5 sec x + 10
    2 (sec^2 x -1) = 5 sec x + 10
    2 (sec^2 x -1) - 5 sec x - 10 = 0
    2 sec^2 x -2 - 5 sec x - 10 = 0
    (2 sec x + 3)(sec x - 4) = 0
    sec x = 3/2 or sec x = 4
    sec x^2 = (3/2)^2 or sec x^2 = 4^2
    sec x^2 = 9/4 or sec x^2 = 16
    (sec x^2) - 1 = 9/4 -1 or (sec x^2) - 1 = 16 - 1
    tan x^2 = 5/4 or tan x^2 = 15
    tan x = √(5/4) or tan x = √15, -√15
    tan x = (√5)/2, -(√5)/2 or tan x = √15, -√15
    For x to be obtuse, tan x must be in 2nd quadrant => negative
    tan x = -(√5)/2 or tan x = -√15 (rejected)

    My question: why is tan x = - √15 rejected?
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  2. #2
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    Quote Originally Posted by ppppp77 View Post


    My question: why is tan x = - √15 rejected?

    because...

    Quote Originally Posted by ppppp77 View Post

    obtuse : 90deg < x< 180 deg
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  3. #3
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    oh my, pickslides, you are awesome!!
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  4. #4
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    but can i check, cause my understanding is as below:
    tan x = - √15, so x is in 2nd or 4th quadrant
    acute angle = 75.5
    x = 180 - 75.5 = 104.4 which qualifies as answer because 90deg < x< 180 deg??
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  5. #5
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    Hello ppppp77
    Quote Originally Posted by ppppp77 View Post
    but can i check, cause my understanding is as below:
    tan x = - √15, so x is in 2nd or 4th quadrant
    acute angle = 75.5
    x = 180 - 75.5 = 104.4 which qualifies as answer because 90deg < x< 180 deg??
    In your working you had (correctly):
    \tan^2x = 15
    and you then said (correctly):
    \tan x = \pm \sqrt{15}
    You then (correctly) rejected \tan x = \sqrt{15} because x is obtuse and therefore \tan x< 0.

    But does that mean that \tan x = -\sqrt{15} is necessarily correct? No, before you can say that, there's one more test to apply, and that is to see whether this value satisfies the original equation; which was:
    2\tan^2x = 5 \sec x +10,\; 90^o<x<180^o
    Now the LHS is 2\times15=30. But what about \sec x on the RHS? Well, using \sec^2x = 1+ \tan^2 x, we get:
    \sec^2x=16

    \Rightarrow \sec x = \pm4
    and, of course, for 90^o<x<180^o, we must take the negative sign, giving
    \sec x = -4
    and the RHS then is -20+10 = -10, not +20+10= 30 (which is what we wanted).

    So \tan x = -\sqrt{15} doesn't work either.

    Tricky little beast, isn't it?

    Grandad
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