# (1-cos A)(1+sec A)=tan A sin A. Please answer this question faster

Sujit Kumar
111 Points
4 years ago
I am assuming the question is : Prove that SinATanA=(1-CosA)(1+SecA)

$(1-CosA)(1+SecA)=TanASinA$
$Transposing \ (1-CosA) \ to \ RHS$
$=>(TanA)(SinA)(\frac{1}{1-CosA})=(1+SecA)$__________(1)
Taking only the LHS from equation (1)
$=>(\frac{SinA}{CosA})(SinA)(\frac{1}{1-CosA})$
$=>(Sin^2A)(\frac{1}{CosA-Cos^2A})$
$=>\frac{Sin^2A}{CosA-Cos^2A}$
$(Note: \ Sin^2x=1-Cos^2x)$
$=>\frac{1-Cos^2A}{CosA-Cos^2A}$
$(Note: \ a^2-b^2=(a+b)(a-b))$
$=>\frac{(1+CosA)(1-CosA)}{CosA(1-CosA)}$
$=>\frac{(1+CosA)}{CosA}$
$=>\frac{1}{CosA}+1$
$=>SecA+1=RHS \ of \ equation \ (1)$
$Hence \ Proved \ !$