Question

in a random triangle, prove that : tan A - cot B - cot C = tanA.cotB.cotC

Answer 1

Hence it is proved that, tan A - cot B - cot C = tanA.cotB.cotC

Given,

tan A - cot B - cot C = tanA.cotB.cotC

tan A - 1/tan B - 1/tan C = tanA.cotB.cotC

tanAtanBtanC-tanC-tanB / tanBtanC = tanA / tanBtanC

tanAtanBtanC-tanC-tanB = tanA

tanAtanBtanC = tanA + tanB + tanC

tanAtanBtanC - tanA = tanB + tanC

tanA ( tanBtanC - 1) = tanB + tanC

tanA = (  tanB + tanC ) / ( tanBtanC - 1)

tanA = - (tanB+tanC)/(1-tanBtanC)

tanA = - tan(B-C) ...........(1)

but, we have, B-C = π-A

tan(π-∅) = - tan∅

∴ tan(B-C) = tan(π-A) = -tanA

substituting the above value in eqn (1), we get,

tanA = - (-tanA)

tanA = tanA.

Hence proved.

Answer 2

 tan A - cot B - cot C  =  tan A . cot B . cot C   , proved

Step-by-step explanation:

Given as :

In a Triangle A B C

To Prove : tan A - cot B - cot C = tan A . cot B . cot C

From Left hand side

tan A - cot B - cot C

= tan A - $\dfrac{1}{tan B}$  -  $\dfrac{1}{tan C}$                                                  ( ∵ $tan\Theta = \dfrac{1}{cot\Theta }$  )

Taking LCM

= $\dfrac{tan A tan B tan C -tan C-tan B }{tan B tan C}$               ..........1

∵ In any Triangle, A + B + C = π

Or, A + B  = π - C

tan ( A + B ) = tan ( π - C )

Or, $\dfrac{tan A + tan B}{1-tan A tan B}$ = - tan C

Or, tan A + tan B = - tan C + tan A tan B tan C

Or, tan A tan B tan C =  tan A + tan B + tan C            .......2

So, from eq 1 and e 2

= $\dfrac{tan A+ tan B+ tan C -tan C-tan B }{tan B tan C}$

= $\dfrac{tan A}{tan BtanC}$

= tan A . cot B . cot C                                  ( ∵ $tan\Theta = \dfrac{1}{cot\Theta }$  )

So,  tan A - cot B - cot C  =  tan A . cot B . cot C      

i.e  Left hand side = Right hand side

Hence,  tan A - cot B - cot C  =  tan A . cot B . cot C   , proved  Answer