Sintheta/cottheta+cosectheta =2 + sintheta/cottheta-cosectheta
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Answer:
LHS = sinA/(cotA+cosecA)
= sinA/[(cosA/sinA)+(1/sinA)]
= sinA/[(cosA+1)/sinA]
= sin²A/(1+cosA)
= (1-cos²A)/(1+cosA)
/* sin²A = 1 - cos²A */
= [(1+cosA)(1-cosA)]/(1+cosA)
/* a²-b² = (a+b)(a-b) */
= 1 - cosA ----(1)
RHS = 2+ [sinA/(cotA-cosecA)]
= 2+sinA/[(cosA/sinA)-(1/sinA)]
= 2+SinA/[(cosA-1)/sinA]
= 2+ [ sin²A/(cosA-1)]
= 2 - ( sin²A)/(1-cosA)
= 2- [ (1-cos²A)/(1-cosA)]
= 2-[(1+cosA)(1-cosA)/(1-cosA)]
= 2 - ( 1+cosA)
= 2-1-cosA
= 1-cosA ----(2)
Form (1) & (2) , we conclude that,
(1) = (2)
LHS = RHS
••••
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