 # Doubt on Trigonometry

Extreme left side is actually 4

If possible, Then plz provide full solution.Thanks

Sir my answer is 12, above equation only possible for a=b=c=d= pi/6 ,

On putting π/6, we get 9√3 , not 3√3 @Raghudevram_Singh

Are you sure

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Ohh sorry i take tan(60°) Anyway , how do u think that 60° will satisfy bro , i mean by hit and trial?

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Now answer is definitely 12,it seem that your mind in @60 ,well it comes from experience ,also you can get by am GMmay

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Yes @Raghudevram_Singh you are Right.

Actually How we can prove that a=b=c=d=60^\circ is only solution.

May @Hari_Shankar bhatt sir will help us. Thanks

Sir it 30 not v60

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Ohh yes it is a=b=c=d=30^\circ

We have \cos A \cos B \cos C \cos D (\tan A +\tan B+ \tan C +\tan D)

=\cos D [\cos A \cos B \cos C(\tan A +\tan B+ \tan C)]+ \sin D \cos A \cos B \cos C

=cos D [\sin (A+B+C)+\sin A \sin B \sin C]+\sin D \cos A \cos B \cos C

\le \sqrt{ [\sin (A+B+C)+\sin A \sin B \sin C]^2+(\cos A \cos B \cos C)^2}

Conisider F(A,B,C) = [\sin (A+B+C)+\sin A \sin B \sin C]^2+(\cos A \cos B \cos C)^2

For a given value of A+B+C, by Jensen's inequality both \sin A \sin B \sin C and \cos A \cos B \cos C attain their maximum when A=B=C= \theta

Then the expression = (\sin 3 \theta + \sin^3 \theta)^2+\cos^6 \theta

=8 \sin^6 \theta-15 \sin^4 \theta+6 \sin^2 \theta+1=8t^3-15t^2+6t+1=P(t) where t=\sin^2 \theta

We see that P(t) attains its maximum when t=\dfrac{1}{4} or when \sin \theta = \dfrac{1}{2} or when A=B=C = \dfrac{\pi}{6}

Furthermore the original expression attains maximum when

\dfrac{\cos D}{ [\sin (A+B+C)+\sin A \sin B \sin C]} = \dfrac{\sin D}{\cos A \cos B \cos C} or when \tan D = \dfrac{1}{\sqrt 3} which happily for us is at D=\dfrac{\pi}{6}

Thus the maximum value of the expression 4\cos A \cos B \cos C \cos D (\tan A +\tan B+ \tan C +\tan D) is attained at A=B=C=D=\dfrac{\pi}{6} and equals 3 \sqrt 3

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Thanks so much Bhatt Sir, Sir plz have a look on

Sir i did now by cauchy inequality (vector form)..