IF SecA+TanA=p P.T SinA=p2[Square]-1/p2[Square]+1

Asked by chethan U | 23rd Jul, 2012, 06:32: PM

Expert Answer:

We know that sec2 A - tan2 A =1 and it is given that
sec A + tan A =p

dividing equation 1 with 2, we get

sec A - tan A = 1/p
Also sec A + tan A = p
Adding and subtracting these equations, we get
sec A = (p + 1/p )/2
& tan A = (p-1/p)/2
Thus, sin A = (sin A/ cos A) / (1/cos A) = tan A / sec A =(p2 - 1) / (p2+ 1)
 

Answered by  | 23rd Jul, 2012, 10:42: PM

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