Answer :
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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