If sinθ = a/b, show that


Consider LHS,


LHS = secθ + tanθ =


= (1)


We have, sinθ = (ak)/(bk) = BC/AB (For some value of k)



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


∴AB2 = BC2 + AC2


= (bk)2 = (ak)2 + AC2


= AC2 = b2k2 - a2k2


→ AC =


∴ cosθ = AC/AB = =


∴ from(1)


LHS = =


=


=


= RHS


HENCE PROVED


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