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sabika13

  • 2 years ago

How do I solve this using trig identities: tan(pi/4 + x) +tan(pi/4 - x) =2sec2x

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  1. sabika13
    • 2 years ago
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    i tried left side: 1+tanx+1-tanx =2 :S

  2. nubeer
    • 2 years ago
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    have you studied double angle identities?

  3. sabika13
    • 2 years ago
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    well i have the formulas for it

  4. nubeer
    • 2 years ago
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    ok good tell me formula for tan(a+b) and tan (a-b) remember a and b are angles.

  5. sabika13
    • 2 years ago
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    tan(a+/-b)=(tana+/-tanb)/(1+/-tanatanb)

  6. nubeer
    • 2 years ago
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    good.. now try to use this formula in your question.. it would be some thing like this for tan(pi/4 +x) = [tan(pi/4) +tanx]/(1-tan(pi/4)tanx)

  7. sabika13
    • 2 years ago
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    and then do i conjugate?

  8. nubeer
    • 2 years ago
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    yes.. basically tan(pi/4) = 1 so just start solving as pi/4 will be eliminated.. the rest would be LCM and simple algebra.

  9. nubeer
    • 2 years ago
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    i think it should work.. as where 2 angles are involved we use double angle identities to make stuff easy for us.. so this should work out.

  10. sabika13
    • 2 years ago
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    i ended up with tanx

  11. sabika13
    • 2 years ago
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    actually it was 1+tan^2x/1+tanx

  12. nubeer
    • 2 years ago
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    seriously.. hmm i will solve it and will see.. i have to go right now but will be back in a while just post the working u did.. if i can do i will post back..

  13. nubeer
    • 2 years ago
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    hmm i dont think so you did right.. as far as i know there should be 1-tan^2x in the denominator.

  14. sabika13
    • 2 years ago
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    i got (1+tanx)(1-tanx)in deno

  15. nubeer
    • 2 years ago
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    yes.. multiplying it would give (1-tan^2x)

  16. sabika13
    • 2 years ago
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    okaylet me show what i did: tanpi/4 + tanx)^2/(1-tanpi/4tanx)(1+tanpi/4tanx) + tanpi/4-tanx/1+tanpi/4+tanx

  17. sabika13
    • 2 years ago
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    right?

  18. sabika13
    • 2 years ago
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    (tanpi/4+tanx)^2 + (tanpi/4-tanx) / (1-tanpi/4tanx)(1+tanpi/4tanx)

  19. sabika13
    • 2 years ago
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    and then tanpi/4 =1

  20. sabika13
    • 2 years ago
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    2+tan^2x-tanx / 1-tan^2x

  21. sabika13
    • 2 years ago
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    ru there?

  22. nubeer
    • 2 years ago
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    yes i am.. just solving for you..

  23. nubeer
    • 2 years ago
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    ok well fault i found in your answer is in frist step. tanpi/4 + tanx)^2 how you got the square here?

  24. sabika13
    • 2 years ago
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    i multiplied (tan pi/4 +tanx) to both top and bottom

  25. sabika13
    • 2 years ago
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    it will become a common deno

  26. nubeer
    • 2 years ago
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    ok i got easy way.. try wolfram.. http://www.wolframalpha.com/input/?i=tan%28pi%2F4+%2B+x%29+%2Btan%28pi%2F4+-+x%29+%3D2sec%282x%29&dataset=&equal=Submit

  27. nubeer
    • 2 years ago
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    tell me if this helps you.

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