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Julie-0606

  • 3 years ago

Trigonometry Question

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  1. Julie-0606
    • 3 years ago
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  2. DHASHNI
    • 3 years ago
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    let teta be 'x' and phi be 'y' tan(x+y) = a/b \[\frac{ \tan x +\tan y }{ 1- \tan x tany} = \frac{ a }{ b }\]

  3. DHASHNI
    • 3 years ago
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    |dw:1363532494603:dw|

  4. DHASHNI
    • 3 years ago
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    tan |dw:1363532542768:dw|

  5. Julie-0606
    • 3 years ago
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    It gives you answer choices. I dont know if this helps

  6. Julie-0606
    • 3 years ago
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  7. DHASHNI
    • 3 years ago
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    yeah wait

  8. Julie-0606
    • 3 years ago
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    ok

  9. DHASHNI
    • 3 years ago
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    cos x = CD/a cos y =CD/b (cos x/cos y )=b/a ------------------------->i =>(cosx)(a/b) =cos y \[=>(cosx)(a/b) = \sqrt{1-\sin^2 y}\]

  10. DHASHNI
    • 3 years ago
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    \[\sin y =\sqrt{1-\frac{ a^2 }{ b^2 }\cos^2x}\]

  11. DHASHNI
    • 3 years ago
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    ------ii

  12. DHASHNI
    • 3 years ago
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    now multiplying sin y in equation i \[ \tan y \cos x =(\frac{ b }{ a})\sin y\]

  13. DHASHNI
    • 3 years ago
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    substituting equation ii in this \[\tan y=\frac{ b }{ a }\frac{ \sqrt{1- \frac{ a^2 }{ b^2} \cos^2 x} }{ cosx }\]

  14. DHASHNI
    • 3 years ago
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    \[=> \tan y = \frac{ \sqrt{{b^2-a^2 \cos^2x}} }{ acos x }\]

  15. DHASHNI
    • 3 years ago
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    option 6 is the answer

  16. Julie-0606
    • 3 years ago
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    Thank you :)

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