InterviewSolution
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InterviewSolution
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In a `Delta ABC`, if `cosA cos B cos C= (sqrt3-1)/(8) and sin A sin B sin C= (3+ sqrt3)/(8)`, then The value of `tan A + tan B + tan C` isA. `(3+ sqrt3)/(sqrt3-1)`B. `(sqrt3 +4)/(sqrt3-1)`C. `(6-sqrt3)/(sqrt3-1)`D. `(sqrt3+ sqrt2)/(sqrt3-1)` |
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Answer» Correct Answer - A `because cos A cos B cos C = (sqrt3 -1)/(8)` `sin A sin B sin C = (3+sqrt3)/(8)` `therefore tanA tan B tan C = (3+ sqrt3)/(sqrt3-1)" " `...(1) `therefore tan A+ tanB + tan C = tan A tan B tan C ` `" " = (3 + sqrt3)/(sqrt3 -1) " "` ...(2) Now `A + B +C= pi` `therefore cos(A+B +C) = -1 ` `therefore cos A cos B cos C [1 -sum tan A tan B] =-1` `therefore sum tanA tan B = 5 + 4sqrt3" "` ...(3) Form (1), (2) and (3), we get `tanA, tan B, tanC` are roots of `x^(3)- ((3+sqrt3)/( sqrt3-1)) x^(2) + (5+4sqrt3)x - ((3+sqrt3))/(sqrt3-1)=0` or `x^(3) - (2+ sqrt3)sqrt3 x^(2) + (5+4sqrt3)x - (2+ sqrt3) sqrt3 =0` or ` x^(3) - (3 + 2 sqrt3)x^(2) + (5+ 4sqrt3)x - (3+ 2sqrt3)=0` or `(x-1)(x-sqrt3)(x-(2+sqrt3)) =0` `therefore tanA =1, tan B = sqrt3, tan C = 2 + sqrt3` |
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