If tan θ, 2 tan θ + 2, 3 tan θ + 3 are in G.P. then the value of 7−5 cot θ9−4 √sec2 θ−1 is
(2 tan θ+2)2=tan θ(3 tan θ+3)⇒ tan2 θ+5 tan θ+4=0⇒ tan θ=−1,−4 but tan θ = – 4 is only satisfies above condition