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In a triangle ABC if the sides a,b,c are in AP and 2/1!.9!+2/3!.7!+1/5!.5!=8a/(2b)!,then the maximum value of tanA.tanB is equal to
A)1/2 B)1/3 C) 1/4 D) 1/5
2/1!.9! + 2/3!.7! + 1/5!.5! = 23a/(2b)!
simply solving above gives
1/(4*6!) *(1/5 +13/63)
512/10! = 23a/(2b)!
so a=3 , b =5 , c=7
area of triangle =15*31/2/4
sin A = 2*area /bc = 3*31/2/14
sin B = 2*area /ac = 5*31/2/14
Cos A = 13/14 , cos B = 11/14
tan A.TanB = 45/143
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regards
Ramesh
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