If $$\frac{{{{\sin }^2}\phi - 3\sin \phi + 2}}{{{{\cos }^2}\phi }} = 1,$$     where 0° < $$\phi $$ < 90° then what is the value of (cos2$$\phi $$ - sin3$$\phi $$ + cosec2$$\phi $$)?

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{{\sin }^2}\phi - 3\sin \phi + 2}}{{{{\cos }^2}\phi }} = 1 \cr & \Rightarrow {\sin ^2}\phi - 3\sin \phi + 2 = {\cos ^2}\phi \cr & \Rightarrow {\sin ^2}\phi - 3\sin \phi + 2 = 1 - {\sin ^2}\phi \cr & \Rightarrow 2{\sin ^2}\phi - 3\sin \phi + 1 = 0 \cr & \Rightarrow 2{\sin ^2}\phi - 2\sin \phi - \sin \phi + 1 = 0 \cr & \Rightarrow 2\sin \phi \left( {\sin \phi - 1} \right) - 1\left( {\sin \phi - 1} \right) = 0 \cr & \Rightarrow \left( {\sin \phi - 1} \right)\left( {2\sin \phi - 1} \right) = 0 \cr & \therefore \sin \phi = 1\,\,\left( {{\text{not valid}}} \right) \cr & {\text{and }}\sin \phi = \frac{1}{2} \cr & \Rightarrow \sin \phi = \sin {30^ \circ } \cr & \therefore \phi = {30^ \circ } \cr & \therefore \cos 2\phi - \sin 3\phi + {\text{cosec }}2\phi \cr & = \cos {60^ \circ } - \sin {90^ \circ } + {\text{cosec }}{60^ \circ } \cr & = \frac{1}{2} - 1 + \frac{2}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} - \frac{1}{2} \cr & = \frac{{\left( {4 - \sqrt 3 } \right)}}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{4\sqrt 3 - 3}}{6} \cr & \cr & {\bf{Alternative:}} \cr & {\text{Put }}\phi = {30^ \circ } \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \frac{{{{\sin }^2}{{30}^ \circ } - 3\sin {{30}^ \circ } + 2}}{{{{\cos }^2}{{30}^ \circ }}} \cr & = \frac{{\frac{1}{4} - \frac{3}{2} + 2}}{{\frac{3}{4}}} \cr & = \frac{{1 - 6 + 8}}{3} \cr & = \frac{3}{3} \cr & = 1 = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \therefore \cos 2\phi - \sin 3\phi + {\text{cosec }}2\phi \cr & = \cos {60^ \circ } - \sin {90^ \circ } + {\text{cosec }}{60^ \circ } \cr & = \frac{1}{2} - 1 + \frac{2}{{\sqrt 3 }} \cr & = - \frac{1}{2} + \frac{2}{{\sqrt 3 }} \cr & = \frac{{\left( { - \sqrt 3 + 4} \right)}}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{ - 3 + 4\sqrt 3 }}{6} \cr} $$