In a Δ ABC, if cos A. cos B. cos C = 
sin A. sin B. sin C = 
, then
(i) Value of tan A + tan B + tan C is
Text Solution
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Ans.
(i)
Sol. cos A cos B cos C = 
sin A sin B sin C = 
∴ tan A tan B tan C = 
… (i)
Now tan A + tan B + tan C = tan A tan B tan C
= 
… (ii)
Now A + B + C = π
cos (A + B + C) = –1
cos A cos B cos C [1 – Σ tan A tan B] = –1
[1– Σ tan A tan B] –1
⇒ Σ tan A tan B = 5 + 4 
… (iii)
from (1), (2) & (3)
tan A, tan B , tan C are roots of
x 3 – 
x 2 + (5 + 4 
) x – 
= 0
x 3 – (2 + 
) 
x 2 + (5 + 4 
) x – (2 + 
) 
=0
1x 3 – (3 + 2 
)x 2 + (5 + 4 
)x –
(3 + 2 
) = 0
(x –1) (x – 
) (x – (2 + 
)) = 0 ∴ tan A = 1, tan B = 
, tan C = 2 + 
(ii)
Sol. cos A cos B cos C = 
sin A sin B sin C = 
∴ tan A tan B tan C = 
… (i)
Now tan A + tan B + tan C = tan A tan B tan C
= 
… (ii)
Now A + B + C = π
cos (A + B + C) = –1
cos A cos B cos C [1 – Σ tan A tan B] = –1
[1– Σ tan A tan B] –1
⇒ Σ tan A tan B = 5 + 4 
… (iii)
from (1), (2) & (3)
tan A, tan B , tan C are roots of
x 3 – 
x 2 + (5 + 4 
) x – 
= 0
x 3 – (2 + 
) 
x 2 + (5 + 4 
) x – (2 + 
) 
=0
x 3 – (3 + 2 
)x 2 + (5 + 4 
)x – (3 + 2 
) = 0
(x –1) (x – 
) (x – (2 + 
)) = 0
∴ tan A = 1, tan B = 
, tan C = 2 + 
(iii)
Sol. cos A cos B cos C = 
sin A sin B sin C = 
∴ tan A tan B tan C = 
… (i)
Now tan A + tan B + tan C = tan A tan B tan C
= 
… (ii)
Now A + B + C = π
cos (A + B + C) = –1
cos A cos B cos C [1 – Σ tan A tan B] = –1
[1– Σ tan A tan B] –1
⇒ Σ tan A tan B = 5 + 4 
… (iii)
from (1), (2) & (3)
tan A, tan B , tan C are roots of
x 3 – 
x 2 + (5 + 4 
) x – 
= 0
x 3 – (2 + 
) 
x 2 + (5 + 4 
) x – (2 + 
) 
=0
x 3 – (3 + 2 
)x 2 + (5 + 4 
)x – (3 + 2 
) = 0
(x –1) (x – 
) (x – (2 + 
)) = 0
∴ tan A = 1, tan B = 
, tan C = 2 + 
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