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

DHASHNIBest ResponseYou've already chosen the best response.0
tan dw:1363532542768:dw
 one year ago

Julie0606Best ResponseYou've already chosen the best response.0
It gives you answer choices. I dont know if this helps
 one year ago

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

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

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

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

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

DHASHNIBest ResponseYou've already chosen the best response.0
option 6 is the answer
 one year ago
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