If secθ = 17/8, verify that

We have, secθ = (17k)/(8k) = AB/AC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


AB2 = BC2 + AC2


= (17k)2 = BC2 + (8k)2


= 289k2 = BC2 + 64k2


= BC2 = 225k2


BC = 15k


sinθ = BC/AB = (15k)/(17k) = 15/17


cosθ = AC/AB = (8k)/(17k) = 8/17


tanθ = BC/AC = sinθ /cosθ = 15/8


consider the LHS =


=


=


=


= ( - 33)/( - 611)


= 33/611


Now consider RHS =


=


=


= ( - 33)/( - 611)


= 33/611


RHS = LHS


HENCE PROVED


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