Let θ , ϕ ∈ [ 0 , 2 π ] satisfy 2 cos θ ( 1 − sin ϕ ) = sin 2 θ ( tan ( θ /2 ) + cot ( θ /2 )) cos ϕ − 1 , tan ( 2 π − θ ) 0 , and − 1

Question

Let θ , ϕ ∈ [ 0 , 2 π ] satisfy 2 cos θ ( 1 − sin ϕ ) = sin 2 θ ( tan ( θ /2 ) + cot ( θ /2 )) cos ϕ − 1 , tan ( 2 π − θ ) > 0 , and − 1 < sin θ < − 3 ​ /2 . Then ϕ cannot satisfy

TestHub · Accepted Answer

The θ conditions imply 3 π /2 < θ < 5 π /3 . The equation becomes 2 sin ( θ + ϕ ) = 2 cos θ + 1 , whose right side is between 1 and 2. Thus θ + ϕ ∈ ( 13 π /6 , 17 π /6 ) in the admissible range, restricting ϕ to ( π /2 , 4 π /3 ) (with a narrower actual subset). None of A, C, D is possible.

Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

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