Question

(i) sin A tan A / 1-cos A = 1+sec A
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Answer 1

$\mathfrak{\huge{\green{\underline{\underline{Answer :}}}}}$

LHS = Sin A.Tan A.(1/1-Cos A)

=Sin A. (SinA/CosA).(1/1-CosA)

=(Sin^2A)/(CosA.(1-CosA)

=(1-Cos^2A)/( CosA.(1-CosA)

=(1+CosA)(1-CosA)/ CosA.(1-CosA)

=(1/cosA)+(CosA/CosA)

=SecA+1=RHS

Hence Proved

Answer 2

Answer:

$(sinA*sinA/cosA) /(1-cosA)$

$sin^2A/(cosA (1-cosA ))$

$(1-cos^2A)/ (cosA (1-cosA))$

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

$(1+cosA)/cosA$

$(1/cosA)+(cosA/cosA)$

$secA+1$

$1+secA proved$