Question

prove that : cot A + cosA÷ cot A - cos A = cosecA + 1÷ cosecA - 1

Answer 1

cotA -cosA)/(cotA + cosA) = (cosecA-1)/(cosecA+1)


(cotA -cosA)/(cotA + cosA)
((cosA/sinA)-cosA)/((cosA/sinA)-cosA) since cotA=cosA/sinA

taking cosA common from numretor nd denominatr nd cancelling it we get
((1/sinA)-1)/((1/sinA)+1) since 1/sinA=cosecA
therfore
(cosecA-1)/(cosecA+1)

Answer 2

Answer:

cotA -cosA)/(cotA + cosA) = (cosecA-1)/(cosecA+1)

(cotA -cosA)/(cotA + cosA)

((cosA/sinA)-cosA)/((cosA/sinA)-cosA) since cotA=cosA/sinA

taking cosA common from numretor nd denominatr nd cancelling it we get

((1/sinA)-1)/((1/sinA)+1) since 1/sinA=cosecA

therfore

(cosecA-1)/(cosecA+cosecA)

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