In `triangleABC,A+B+C=pi`,show that`cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)`

If A+B+C=180^(@) , then the value of cot(A/2)+cot(B/2)+cot(C/2) will be .............. A) 2 cot ""A/2 cot ""B/2 cot""C/2 B) 4 cot ""A/2 cot ""B/2 cot ""C/2 C) cot ""A/2 cot ""B/2 cot ""C/2 D) 8 cot ""A/2 cot ""B/2 cot ""C/2

If a, b, c are in A.P then cot(A/2),cot(B/2),cot(C/2) are in

(1 - cos^(2)theta)cot^(2)theta

cosec A- 2cot 2A cos A=

Using tables evaluate the following (frac(cot (pi/4) cot (pi/3)+1)(cot (pi/4)-cot (pi/3)))

In DeltaABC,cot.(A+B)/(2).tan.(A-B)/(2) =

In /_\ABC , prove that cot(frac{A}{2})+cot(frac{B}{2})+cot(frac{C}{2}) = frac{a+b+c}{b+c-a}cot(frac{A}{2})

In DeltaABC,(cot((A)/(2))+cot((B)/(2)))(asin^(2)((B)/(2))+bsin^(2)((A)/(2))) =

Prove the following: cos(3pi/2+x)cos(2pi+x)[cot(3pi/2-x)+cot(2pi+x)]=1

If cosA=mcosB , then ............. A) cot ((A+B)/(2)) =(m+1)/(m-1) tan ((B-A)/(2)) B) tan ((A+B)/(2)) =(m+1)/(m-1) cot ((B-1)/(2)) C) cot ((A+B)/(2)) =(m+1)/(m-1) cot ((B-A)/(2)) D) None of these