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Grade: 10
        
If x+y+z=pi
Prove:
Cotx/2 + coty/2 + cotx/2=cotx/2*coty/2*cotz/2
 
2 months ago

Answers : (2)

Aditya Gupta
1673 Points
							
we know the identity:
cot(a+b)= (cota*cotb – 1)/(cota + cotb)
so that cota + cotb= (cota*cotb – 1)/cot(a+b)
in above eqn put a= x/2 and b= y/2 you get
cotx/2 + coty/2= (cot(x/2)*cot(y/2) – 1)/cot[(x+y)/2]
now, x/2 + y/2 + z/2= pi/2
so cot[(x+y)/2]= cot(pi/2 – z/2)= tan(z/2)
so cotx/2 + coty/2= (cot(x/2)*cot(y/2) – 1)/tan(z/2)= (cot(x/2)*cot(y/2) – 1)*cot(z/2)= cotx/2*coty/2*cotz/2 – cotz/2
or cotx/2 + coty/2 + cotx/2=cotx/2*coty/2*cotz/2
kindly approve :)
2 months ago
PRATEEK PANDEY
15 Points
							
x+y=180-z
or, x/2+y/2=90-z/2
or, cot(x/2+y/2)=cot(90-z/2)
or, (cotx/2*coty/2 -1)/(cotx/2+coty/2)=tanz/2
 or, (cotx/2*coty/2 -1)/(cotx/2+coty/2)=1/cotz/2
or, cotx/2*coty/2*cotz/2-cotz/2=cotx/2+coty/2
or,  cotx/2*coty/2*cotz/2 =cotx/2+coty/2+cotz/2          PROVED..
 
 
2 months ago
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