Question

1+secA/secA=sin²A/1-cosA​

Answer 1

(1+secA)/secA=sin²A/(1-cosA)

Taking L.H.S

(1+secA)/secA

1/secA+secA/secA

cosA+1

Multiplying and dividing by 1-cosA

((1+cosA)(1-cosA))/(1-cosA)

(1-cos²A)/(1-cosA)

But 1-cos²A= sin²A

sin²A/(1-cosA)

L.H.S=R.H.S

Hence Proved

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