2 Answers
11Ans ) tan 2 θ + sec θ = λ
Therefore, sec 2 θ + sec θ - (λ + 1) = 0
sec θ = -1 ± √( 4 λ + 5 )2
Now, for real sec θ , 4 λ + 5 ≥ 0 or λ ≥ -5/4 ...........(i)
CASE I :-
sec θ ≥ 1
Therefore, -1 ± √( 4 λ + 5 )2 ≥ 1
4 λ +5 ≥ 9
or λ ≥ 1 ....................(ii)
CASE II :-
sec θ ≤ -1
or, -1 ± √( 4 λ + 5 )2 ≤ - 1
or 4 λ +5 > 1
or λ ≥ - 1 ...................(iii)
From (i) , (ii) and (iii), we get
λ belonging to [-1 , ∞ )
Thus, answer is (d) [1]
62a simpler solution would have been that
the question is same as finding the range of the LHS
which is same as tan2θ+sec θ = sec2θ+sec θ-1
which is not very difficult.. given that sec lies outside (-1,1)
