Question

show that cot A. cot B + cot B .cot C +cot C. cot A =1

Answer 1

Hey !!!

Let A , B and C are angle of triagle

Hence, sum of all angle of triangle= 180 °

Since,

<A + <B + <C = 180 °

<A + <B = 180° - <C

take cot on both side
we get ,

cot ( A + B ) = cot ( 180° - C)

cot ( A + B ) = - cotC

we know that ,

cot ( A + B ) = cotA * cotB - 1 / cotA + cotB

since, using this formula

cotA* cotB - 1 / cotA + cotB = - cotC

cotA* cotB - 1 = - ( cotA* cotC ) -( cotB * cotC)

cotA * cotB + cotA* cotC + cotB * cotC = 1

since, here prooved ♻

cotA* cotB + cotB* cotC + cotC * cotA = 1

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$\bf{hope \: it \: \: helps \: you}$

$\underline{rajukumar111}$

Answer 2

Step-by-step explanation:

In ∆ ABC , A+B+C = π.

or, (A + B ) = (π - C ).

or, cot (A+B) = cot (π -C).

or, ( cotA.cotB -1)/(cotB + cotA) = - cotC.

or , cotA.cotB -1 = - cotB.cotC - cotC.cotA.

or , cotA.cotB +cotB.cotC + cotC.cotA. = 1

Proved

Hope it helps.

Thank You.