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Grade: 10
        
cosecA + cosec2A + cosec4A + cosec 8A …. n terms =
4 years ago

Answers : (4)

Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
							If its


Now its doesn't seem to solve into anything.


Arun Kumar
4 years ago
Sanip Panta
25 Points
							The answer of this question is: CotA-cotnA    °n is the angle of last term as cosecnA. For example:CosecA+cosec2A+cosec4A+cosec8A=cotA-cot8A
						
one year ago
Sanip Panta
25 Points
							The answer was a bit mistake. Actually it was cotA/2 -  cotnA #bA is the angle of last term.  For example :CosecA+cosec2A +cosec4A +cosec8A +cosec16A = cotA/2 - cot16A
						
one year ago
Shree Ram Gupta
17 Points
							= cotA - cot8A= cotA - cos8A/sin8A= cotA - 2cos^2(4A) - 1/2sin4Acos4A= cotA - 2cos^2(4A)/2sin4Acos4A + 1/2sin4Acos4A= cotA - cos4A/sin4A + 1/sin8A= cotA - 2cos^2(2A) - 1/2sin2Acos2A + cosec8A= cotA - 2cos^2(A)/2sin2Acos2A + 1/2sin2Acos2A + cosec8A= cotA - cos2A/sin2A+ 1/sin4A+cosec8AcotA - 2cos^2(A) - 1/2sinAcosA + cosec4A + cosec8A=cosA/sinA - cosA/sinA + 1/sin2A + cosec4A + cosec8A= cosec2A + cosec4A + cosec8A
						
10 months ago
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