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maheshmeghwal9

Inverse Trigonometric Function Question:- Please Help:)

  • one year ago
  • one year ago

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  1. maheshmeghwal9
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    • one year ago
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  2. maheshmeghwal9
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    Why tanx≠−2 ?????

    • one year ago
  3. maheshmeghwal9
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    Please give reason:)

    • one year ago
  4. maheshmeghwal9
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    @apoorvk & @Callisto & @shivam_bhalla Plz help:)

    • one year ago
  5. Mertsj
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    \[\tan \frac{1}{2}x=\frac{1-\cos x}{\sin x}\]

    • one year ago
  6. maheshmeghwal9
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    What is this?

    • one year ago
  7. Mertsj
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    \[\tan(\frac{1}{2}\cos^{-1} \frac{3}{5})=\frac{1-\frac{3}{5}}{\frac{4}{5}}=\frac{1}{2}\]

    • one year ago
  8. maheshmeghwal9
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    but answer is given "2" in the question.

    • one year ago
  9. FoolForMath
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    \(\frac 12 \) is the correct answer.

    • one year ago
  10. maheshmeghwal9
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    k! thanx to everyone:)

    • one year ago
  11. maheshmeghwal9
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    but in the question; is the solution wrong anywhere? Plz tell:)

    • one year ago
  12. FoolForMath
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    In the componendo dividendo step, the final answer is \(\tan^2 x =\frac 1 4 \)

    • one year ago
  13. shivam_bhalla
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    This is how @Mertsj got that and it is right \[\large \tan x/2 = \frac{\sin x/2 }{\cos x/2} = {\sqrt{1-cosx \over 2} \over \sqrt{1+cosx \over 2}}\] \[\large \tan x/2 = {\sqrt{1-cosx } \over \sqrt{1+cosx}} * \frac{\sqrt{1+cosx}}{\sqrt{1+cosx}}\] \[\large \tan x/2= {sinx \over 1+\cos x }\]

    • one year ago
  14. maheshmeghwal9
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    k!k! I got that thanx a lot:)

    • one year ago
  15. maheshmeghwal9
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    I want to know why answer is not -1/2

    • one year ago
  16. maheshmeghwal9
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    First u see full cmments

    • one year ago
  17. KingGeorge
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    It seems that trying to attack it your way, and ending with a \(\tan^2\), there's no real easy way to see that it can't be -1/2. However, doing it mertsj's method, only outputs the single solution.

    • one year ago
  18. maheshmeghwal9
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    I didn't understand ur 1st line:\

    • one year ago
  19. maheshmeghwal9
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    Plz help:)

    • one year ago
  20. KingGeorge
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    I'm going off of what foolformath said, "In the componendo dividendo step, the final answer is \(\tan^2(x)=\frac{1}{4}\)." I can't see a good reason why you need to exclude the possible answer of -1/2 here. Also, \(\sin(\cos^{-1}(3/5))\) should have two solutions. Namely, 4/5 and -4/5. So you should still get both solutions.

    • one year ago
  21. maheshmeghwal9
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    k! thanx I got it:)

    • one year ago
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