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

Solution:
$$\eqalign{ \tan \theta = \frac{3}{4} \cr {\text{cos2}}\theta = \frac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} \cr {\text{cos2}}\theta = \frac{7}{{25}} \cr} $$

Solution:
$$\eqalign{ \Rightarrow {\text{sin 3A}} = {\text{cos}}\left( {{\text{A}} - {{26}^ \circ }} \right) \cr \Rightarrow {\text{3A + A}} - {\text{2}}{{\text{6}}^ \circ }{\text{ = }}{90^ \circ } \cr \left[ {{\text{If sin A}} = {\text{cos B}}{\text{. then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr \Rightarrow 4{\text{A = 11}}{6^ \circ } \cr \Rightarrow {\text{A}} = {29^ \circ } \cr} $$

Solution:
$$\eqalign{ {\sec ^2}{28^ \circ } - {\cot ^2}{62^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr = {\sec ^2}{28^ \circ } - {\tan ^2}{28^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr = 1 + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( 2 \right)^2} \cr = 1 + \frac{3}{4} + 4 \cr = \frac{{23}}{4} \cr} $$

Solution:
$$\eqalign{ {\text{According to the question ,}} \cr {\text{ se}}{{\text{c}}^4}{\text{A}}\left( {1 - {{\sin }^4}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}} \cr {\text{Put A}} = {45^ \circ } \cr \Rightarrow {\text{ se}}{{\text{c}}^4}{45^ \circ }\left( {1 - {{\sin }^4}{{45}^ \circ }} \right) - 2{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr \Rightarrow 4\left( {1 - \frac{1}{4}} \right) - 2 \cr \Rightarrow 4 \times \frac{3}{4} - {\text{2}} \cr \Rightarrow 3 - 2 \cr \Rightarrow 1 \cr} $$

Solution:
cot10° .cot20° .cot60°.cot70° .cot80°
[ In cot A.cot B if A + B = 90° then cot A.cot B = 1 ]
(cot10° .cot80°) × (cot20°.cot70°) × cot60°
∴ 1 × 1 × $$\frac{1}{{\sqrt 3 }}$$
= $$\frac{1}{{\sqrt 3 }}$$

Solution:
$$\eqalign{ \tan A - \tan B - \tan C = \tan A.\tan B.\tan C \cr \tan A - \tan A.\tan B.\tan C = \tan B + \tan C \cr \tan A = \frac{{\tan B + \tan C}}{{1 - \tan B.\tan C}} \cr \tan A = \tan \left( {B + C} \right) \cr {\text{On comparision,}} \cr A = B + C \cr} $$