Trigonometry Question

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Trigonometry Question

Trigonometry
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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let teta be 'x' and phi be 'y' tan(x+y) = a/b \[\frac{ \tan x +\tan y }{ 1- \tan x tany} = \frac{ a }{ b }\]
|dw:1363532494603:dw|

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Other answers:

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

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