sinA/1+cosA= cosecA- cotA
NeelamG:
let's take RHS cosecA-cotA = 1/sinA - cosA/sinA = (1-cosA)/sinA = (1-cosA)(1+cosA)/sinA(1+cosA) = sinA/(1+cosA)
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Let's take the LHS.
LHS=sinA/(1+cosA)
So now, is we multiply the numerator and denominator bu 1-cosA, we get
sinA*(1-cosA)/(1+cosA)*(1-cosA)
=(sinA-sinA*cosA)/(1-cos²A)
=(sinA-sinA*cosA)/sin²A
=(sinA/sin²A)-((sinA*cosA)/sin²A)
=(1/sinA)-(cosA/sinA)
=cosecA-cotA= RHS
Hence proved.
LHS=sinA/(1+cosA)
So now, is we multiply the numerator and denominator bu 1-cosA, we get
sinA*(1-cosA)/(1+cosA)*(1-cosA)
=(sinA-sinA*cosA)/(1-cos²A)
=(sinA-sinA*cosA)/sin²A
=(sinA/sin²A)-((sinA*cosA)/sin²A)
=(1/sinA)-(cosA/sinA)
=cosecA-cotA= RHS
Hence proved.
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