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Question

State true or false
(b2+c2a2)tanA=(c2+a2b2)tanB=(a2+b2c2)tanC.

A
True
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B
False
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Solution

The correct option is A True
(c2a2+b2)tanA=(C2a2+b2)×sinAcosA

=(c2a2+b2)×akb2+c2a22bc

(c2a2+b2)tanA=2abck ...(1)

(a2b2+c2)tanB=(a2b2+c2)×sinBcosB

=(a2b2+c2)×bk(a2b2+c2)2ac

(a2b2+c2)tanB=2abck ...(2)

(a2+b2c2)tanC

(a2+b2c2)sinCcosC=(a2+b2c2)×ck(a2+b2c2)2ab

=2abck (3)

from (1) (2) and (3)

(c2a2+b2)tanA=(ab2+c2)tanB=(a2+b2c2)tanC
Hence proved.

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