Let θ , φ ∈ [ 0 , 2 π ] be such that 2 cos θ ( 1 − sin φ ) = sin 2 θ × ( tan 2 θ + cot 2 θ ) cos φ − 1 , tan ( 2 π − θ ) 0 and − 1

Question

Let θ , φ ∈ [ 0 , 2 π ] be such that 2 cos θ ( 1 − sin φ ) = sin 2 θ × ( tan 2 θ ​ + cot 2 θ ​ ) cos φ − 1 , tan ( 2 π − θ ) > 0 and − 1 < sin θ < − 2 3 ​ ​ , then φ cannot satisfy

TestHub · Accepted Answer

As tan ( 2 π − θ ) > 0 and − 1 < sin θ < − 2 3 ​ ​ , θ ∈ [ 0 , 2 π ] Hence 2 3 π ​ < θ < 3 5 π ​ Now 2 cos θ ( 1 − sin φ ) = sin 2 θ ( tan 2 θ ​ + cot 2 θ ​ ) cos φ − 1 ⇒ 2 cos θ ( 1 − sin φ ) = 2 sin θ cos φ − 1 ⇒ 2 cos θ + 1 = 2 sin ( θ + φ ) As θ ∈ ( 2 3 π ​ , 3 5 π ​ ) , 1 < 2 sin ( θ + φ ) < 2 As θ + φ ∈ ( 6 π ​ , 6 5 π ​ ) or ( θ + φ ) ∈ ( 6 13 π ​ , 6 17 π ​ ) We have φ ∈ ( − 2 3 π ​ , − 3 2 π ​ ) ∪ ( 3 2 π ​ , 6 7 π ​ )

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