anonymous
  • anonymous
Trigonometry Question
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
let teta be 'x' and phi be 'y' tan(x+y) = a/b \[\frac{ \tan x +\tan y }{ 1- \tan x tany} = \frac{ a }{ b }\]
anonymous
  • anonymous
|dw:1363532494603:dw|

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anonymous
  • anonymous
tan |dw:1363532542768:dw|
anonymous
  • anonymous
It gives you answer choices. I dont know if this helps
anonymous
  • anonymous
anonymous
  • anonymous
yeah wait
anonymous
  • anonymous
ok
anonymous
  • anonymous
cos x = CD/a cos y =CD/b (cos x/cos y )=b/a ------------------------->i =>(cosx)(a/b) =cos y \[=>(cosx)(a/b) = \sqrt{1-\sin^2 y}\]
anonymous
  • anonymous
\[\sin y =\sqrt{1-\frac{ a^2 }{ b^2 }\cos^2x}\]
anonymous
  • anonymous
------ii
anonymous
  • anonymous
now multiplying sin y in equation i \[ \tan y \cos x =(\frac{ b }{ a})\sin y\]
anonymous
  • anonymous
substituting equation ii in this \[\tan y=\frac{ b }{ a }\frac{ \sqrt{1- \frac{ a^2 }{ b^2} \cos^2 x} }{ cosx }\]
anonymous
  • anonymous
\[=> \tan y = \frac{ \sqrt{{b^2-a^2 \cos^2x}} }{ acos x }\]
anonymous
  • anonymous
option 6 is the answer
anonymous
  • anonymous
Thank you :)

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