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Question: In a triangle ABC if cot \(\frac { \mathrm { A } } { 2 }\) cot \(\frac { B } { 2 }\) = c, cot <img...

In a triangle ABC if cot A2\frac { \mathrm { A } } { 2 } cot B2\frac { B } { 2 } = c, cot cot = a and cot cot = b, then + 1s−b\frac { 1 } { s - b } + =

A

–1

B

0

C

1

D

2

Answer

2

Explanation

Solution

cot cot =s(s−a)(s−b)(s−c)×s(s−b)(s−c)(s−a)\sqrt { \frac { s ( s - a ) } { ( s - b ) ( s - c ) } \times \frac { s ( s - b ) } { ( s - c ) ( s - a ) } }= c

⇒ = c ⇒ = similarly = , 1s−b\frac { 1 } { s - b } =

⇒ + 1s−b\frac { 1 } { s - b } + = == 2.