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Question: Let x, y ∈ R such that cosxcosy + 2siny + 2sinxcosy = 3, then find the value of tan²x + 5tan²y....

Let x, y ∈ R such that cosxcosy + 2siny + 2sinxcosy = 3, then find the value of tan²x + 5tan²y.

Answer

8

Explanation

Solution

  1. Rewrite the equation as cosy(cosx+2sinx)+2siny=3\cos y (\cos x+2\sin x) + 2\sin y =3.

  2. Let A=cosx+2sinxA=\cos x+2\sin x; maximum value of Acosy+2sinyA\cos y+2\sin y is A2+4\sqrt{A^2+4} which must equal 3, forcing A2=5A^2=5 and A=5A=\sqrt{5}.

  3. Write cosx+2sinx=5sin(x+α)\cos x+2\sin x = \sqrt{5}\sin(x+\alpha) with sinα=1/5\sin\alpha=1/\sqrt{5}, cosα=2/5\cos\alpha=2/\sqrt{5}.

  4. Then sin(x+α)=1\sin(x+\alpha)=1 leading to tanx=2\tan x=2.

  5. Find tany\tan y from cosy=5/3\cos y=\sqrt{5}/3 and siny=2/3\sin y=2/3 to get tany=2/5\tan y=2/\sqrt{5}.

  6. Finally, compute tan2x+5tan2y=4+4=8\tan^2 x+5\tan^2 y=4+4=8.