1there's a easier method , try using AM-GM inequality . it should help .If you don't get it I'll post the method.
2 Answers
62LHS= 3(cot2θ+tan2θ)-8(cotθ+tanθ)+10
3(cot2θ+tan2θ+2-2)-8(cotθ+tanθ)+10
3(cot2θ+tan2θ+2)-6-8(cotθ+tanθ)+10
3(cot2θ+tan2θ+2) - 8(cotθ+tanθ)+4
3(cotθ+tanθ)2 - 8(cotθ+tanθ)+4
3(cotθ+tanθ)2 - 8(cotθ+tanθ)+4
(cotθ+tanθ)= k
k lies in 2 regions either (-∞,-2] or [2,-∞)
LHS = 3k2-8k+4
if k is in (-∞,-2], this expression is obv greater than 10. bcos all 3 terms will bcom +ve...
LHS has a minimum at 4/3... and increses as k increases.. so the minima will be at k=2 (when we consider k in [2,-∞) )
so at k=2
LHS = 3.4-16+4 = 0 (This is the minimum value it will attain...
1