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Julie0606
 2 years ago
Trigonometry Question
Julie0606
 2 years ago
Trigonometry Question

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DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0let teta be 'x' and phi be 'y' tan(x+y) = a/b \[\frac{ \tan x +\tan y }{ 1 \tan x tany} = \frac{ a }{ b }\]

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0tan dw:1363532542768:dw

Julie0606
 2 years ago
Best ResponseYou've already chosen the best response.0It gives you answer choices. I dont know if this helps

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0cos x = CD/a cos y =CD/b (cos x/cos y )=b/a >i =>(cosx)(a/b) =cos y \[=>(cosx)(a/b) = \sqrt{1\sin^2 y}\]

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sin y =\sqrt{1\frac{ a^2 }{ b^2 }\cos^2x}\]

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0now multiplying sin y in equation i \[ \tan y \cos x =(\frac{ b }{ a})\sin y\]

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0substituting equation ii in this \[\tan y=\frac{ b }{ a }\frac{ \sqrt{1 \frac{ a^2 }{ b^2} \cos^2 x} }{ cosx }\]

DHASHNI
 2 years ago
Best ResponseYou've already chosen the best response.0\[=> \tan y = \frac{ \sqrt{{b^2a^2 \cos^2x}} }{ acos x }\]
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