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`        If x+y+z=piProve:Cotx/2 + coty/2 + cotx/2=cotx/2*coty/2*cotz/2 `
2 months ago

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 getcotx/2 + coty/2= (cot(x/2)*cot(y/2) – 1)/cot[(x+y)/2]now, x/2 + y/2 + z/2= pi/2so 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/2or cotx/2 + coty/2 + cotx/2=cotx/2*coty/2*cotz/2kindly approve :)
```
2 months ago
PRATEEK PANDEY
15 Points
```							x+y=180-zor, x/2+y/2=90-z/2or, 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/2or, cotx/2*coty/2*cotz/2-cotz/2=cotx/2+coty/2or,  cotx/2*coty/2*cotz/2 =cotx/2+coty/2+cotz/2          PROVED..
```
2 months ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions