Question

if A+B+C=180, then the value of(cotB+cot C)(cot C+cotA)(cot A+cot B)is what?

Answer 1

Answer:

Step-by-step explanation:

I think I am one year late but I still hope it helped you.

Answer 2

The value of (cot B + cot C)(cot C + cot A)(cot A + cot B)  = cosec A cosec B cosec C

Step-by-step explanation:

Given: A +B +C = 180

To find: The value of (cot B + cot C)(cot C + cot A)(cot A + cot B)

Solution,

A +B +C = 180

(cot B + cot C)(cot C + cot A)(cot A + cot B)

= $(\frac{Cos B}{Sin B} + \frac{Cos C}{Sin C} ) (\frac{Cos c}{Sin C} + \frac{Cos A}{Sin A} ) + (\frac{Cos A}{Sin A} + (\frac{Cos B}{Sin B} )$

=$(\frac{Cos BSin C+Sin B Cos C}{Sin BSin C} ) (\frac{Cos c Sin A +Sin C Cos A}{Sin CSin A} ) (\frac{Cos ASin B + SinACos B}{Sin ASin B} )$

= $(\frac{sin(B +C)}{sinB Sin C})( \frac{sin(A+C)}{sin C sin A} )(\frac{sin(A +B)}{sin ASin B} )$

= $(\frac{Sin(180-A)}{SinB Sin C})( \frac{sin(180 - B)}{sin C sin A} )(\frac{Sin(180 -C)}{Sin ASin B} )$

Therefore,

$sin(180 -x) = sinx$

⇒$(\frac{sinA}{sin Bsin C}) (\frac{sinB}{sinCsinA} )(\frac{sinC}{sinAsinB} )$

= $\frac{1}{sinAsinBsinC}$

= cosec A cosec B cosec C

Hence,

(cot B + cot C)(cot C + cot A)(cot A + cot B)  = cosec A cosec B cosec C.

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