$$\eqalign{ & {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = {\sin ^2}{70^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & = 1 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow \frac{1}{2} \times \cot {45^ \circ } \cr & \left[ {\frac{{1 + \left( {\sec {{45}^ \circ } + \tan {{45}^ \circ }} \right)}}{{{\text{cosec }}{{45}^ \circ }\left( {\sec {{45}^ \circ } - \tan {{45}^ \circ }} \right)}}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{1 + {{\left( {\sqrt 2 - 1} \right)}^2}}}{{\sqrt 2 \times \left( {\sqrt 2 - 1} \right)}}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{1 + 2 + 1 - 2\sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{4 - 2\sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow \frac{1}{2} \times 2\left[ {\frac{{2 - \sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & = \frac{{\sin \theta .\operatorname{cosec} \theta .\tan \theta .\cot \theta }}{{{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta }} \cr & = \frac{{\sin \theta \times \frac{1}{{\sin \theta }} \times \tan \theta \times \frac{1}{{\tan \theta }}}}{1} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{4co}}{{\text{s}}^2}\theta - 1 = 0 \cr & \Rightarrow {\text{4co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & \Rightarrow \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \theta = {60^ \circ } \cr & \Rightarrow \tan \left( {\theta - {{15}^ \circ }} \right) \cr & \Rightarrow \tan \left( {{{60}^ \circ } - {{15}^ \circ }} \right) \cr & \Rightarrow \tan {45^ \circ } \cr & \Rightarrow 1 \cr} $$

$$\frac{{\sin {{25}^ \circ }.\cos {{65}^ \circ } + \cos {{25}^ \circ }.\sin {{65}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - \cos e{c^2}{{20}^ \circ }}}$$
$$ = \frac{{\sin {{25}^ \circ }.\cos \left( {{{90}^ \circ } - {{25}^ \circ }} \right) + \,\cos {{25}^ \circ }.\,\sin \left( {{{90}^ \circ } - {{25}^ \circ }} \right)}}{{{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right)}}$$
$$\eqalign{ & = \frac{{{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }}}{{{{\tan }^2}{{70}^ \circ } - se{c^2}{{70}^ \circ }}} \cr & = \frac{1}{{ - 1}} \cr & = - 1 \cr} $$

$$\eqalign{ & {\text{cos}}\theta + \sec \theta = \sqrt 3 \cr & {\text{Cubing both sides}} \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3{\text{cos}}\theta \sec \theta \left( {{\text{cos}}\theta + \sec \theta } \right) = 3\sqrt 3 \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3\sqrt 3 = 3\sqrt 3 \cr & {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta = 0 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\sin ^2}{2^ \circ } + {\sin ^2}{4^ \circ } + {\sin ^2}{6^ \circ } + ..... + {\sin ^2}{90^ \circ } \cr & {\text{Number of terms}} \cr & = \frac{{l - a}}{d} + 1 \cr & = \frac{{90 - 2}}{2} + 1 \cr & = 45 \cr & {\text{But }}{\sin ^2}{90^ \circ } = 1 \cr & {\text{So, 22 pairs}} + {\sin ^2}{90^ \circ } \cr & = 22 + 1 \cr & = 23 \cr} $$

$$\eqalign{ & \theta + \phi = \frac{\pi }{2} \cr & \theta + \phi = {90^ \circ }\,......(i) \cr & \sin \theta = \frac{1}{2} \cr & \sin \theta = {\text{sin 3}}{0^ \circ } = \frac{1}{2}\,......(ii) \cr & {\text{Put }}\theta = {30^ \circ }{\text{ in equation (i)}} \cr & {30^ \circ } + \phi = {90^ \circ } \cr & \phi = {60^ \circ } \cr & \sin \phi = \sin {60^ \circ } = \frac{{\sqrt 3 }}{2} \cr} $$

$$\eqalign{ & {\text{The value of }} \cr & {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right) + 2\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) + 12{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta \cr & {\text{Using }}\theta = {0^ \circ } \cr & \because \sin {0^ \circ } = {0^ \circ } \cr & \cos {0^ \circ } = 1 \cr & \Rightarrow 3\left( {0 + {1^4}} \right) + 2\left( {0 + {1^6}} \right) + 12 \times 0 \times 1 \cr & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}}{\text{ + }}\frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ }}}{{{{\sin }^2}{{45}^ \circ }}} - \frac{{{\text{ta}}{{\text{n}}^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{45}^ \circ }}} - \frac{{{{\sin }^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} + \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} - \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}} \cr & \Rightarrow \left( {\frac{1}{2} \times \frac{4}{3}} \right) + \left( {\frac{1}{4} \times \frac{2}{1}} \right) - \left( {\frac{1}{3} \times 1} \right) - \left( {\frac{1}{4} \times \frac{1}{3}} \right) \cr & \Rightarrow \frac{2}{3} + \frac{1}{2} - \frac{1}{3} - \frac{1}{{12}} \cr & \Rightarrow \frac{1}{3} + \frac{1}{2} - \frac{1}{{12}} \cr & \Rightarrow \frac{{4 + 6 - 1}}{{12}} \cr & \Rightarrow \frac{9}{{12}} \cr & \Rightarrow \frac{3}{4} \cr} $$

$$\eqalign{ & {\text{A}} \times {\text{tan}}\left( {\theta + {{150}^ \circ }} \right) = {\text{B}} \times \tan \left( {\theta - {{60}^ \circ }} \right) \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {\theta - {{60}^ \circ }} \right)}}{{\tan \left( {\theta + {{150}^ \circ }} \right)}} \cr & {\text{Put }}\theta = {90^ \circ } \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan \left( {{{90}^ \circ } - {{60}^ \circ }} \right)}}{{\tan \left( {{{90}^ \circ } + {{150}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan \left( {{{180}^ \circ } + {{60}^ \circ }} \right)}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{{\tan {{30}^ \circ }}}{{\tan {{60}^ \circ }}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{1}{3} \cr & {\text{then, }}\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - \frac{4}{2} \cr & \Rightarrow \frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = - 2 \cr & \Rightarrow \frac{{{\text{A}} - {\text{B}}}}{{{\text{A}} + {\text{B}}}} = - \frac{1}{2} \cr & {\text{Put in option (i)}} \cr & - \frac{{\sin {{90}^ \circ }}}{2} = - \frac{1}{2} \cr & {\text{So, option (A) is correct }} \cr} $$

$$\eqalign{ & {\text{tan}}\theta - \cot \theta = 0 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & {\text{tan }}{45^ \circ } - \cot {45^ \circ } = 0 \cr & 1 - 1 = 0 \cr & 0 - 0({\text{matched}}) \cr & So,\theta = {45^ \circ } \cr & \Rightarrow \sin \theta + \cos \theta \cr & \Rightarrow \sin {45^ \circ } + \cos {45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \sqrt 2 \cr} $$

$$\eqalign{ & x\sin {60^ \circ }.\tan {30^ \circ } = \sec {60^ \circ }.\cot {45^ \circ } \cr & {\text{Put values }} \cr & \Rightarrow x.\frac{{\sqrt 3 }}{2}.\frac{1}{{\sqrt 3 }} = 2.1 \cr & \Rightarrow \frac{x}{2} = 2 \cr & \Rightarrow x = 4 \cr} $$

$$\eqalign{ & x\cos \theta - y\sin \theta = \sqrt {{x^2} + {y^2}} \,.....(i) \cr & \frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} = \frac{1}{{{x^2} + {y^2}}}\,.....(ii) \cr & \frac{x}{{\sqrt {{x^2} + {y^2}} }}\cos \theta + \frac{{ - y}}{{\sqrt {{x^2} + {y^2}} }}\sin \theta = 1 \cr & {\text{from equation (i)}} \cr & \Rightarrow \sin \theta = \frac{{ - y}}{{\sqrt {{x^2} + {y^2}} }} \cr & \Rightarrow \cos \theta = \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr & {\text{Put value in equation (ii)}} \cr & \therefore \frac{{{\text{co}}{{\text{s}}^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}} = \frac{1}{{{x^2} + {y^2}}} \cr & \Rightarrow \frac{{{x^2}}}{{\left( {{x^2} + {y^2}} \right){a^2}}} + \frac{{{y^2}}}{{\left( {{x^2} + {y^2}} \right){b^2}}} = \frac{1}{{{x^2} + {y^2}}} \cr & \Rightarrow \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr} $$

$$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = \frac{1}{3}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1{\text{ }}\left( {{\text{Property}}} \right) \cr & \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = \frac{1}{3} \times 1 \cr & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{1}{3} \cr & \therefore \left( {\left( {{a^2} + {b^2}} \right)\left( {{a^2} - {b^2}} \right) = {a^4} - {b^4}} \right) \cr} $$

$$\eqalign{ & {\text{1 + }}\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}} - {\text{se}}{{\text{c}}^2}{27^ \circ }{\text{ + }}\frac{1}{{{{\sin }^2}{{63}^ \circ }}} - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}{63^ \circ } - {\text{se}}{{\text{c}}^2}{27^ \circ } + {\text{cose}}{{\text{c}}^2}{63^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\sec ^2}{27^ \circ } + {\text{se}}{{\text{c}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt 2 }}{\text{sin}}\frac{\pi }{6}{\text{.cos}}\frac{\pi }{4} - \cot \frac{\pi }{3}{\text{.sec}}\frac{\pi }{6}{\text{ + }}\frac{{5\tan \frac{\pi }{4}}}{{12\sin \frac{\pi }{2}}} \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }} \times \frac{2}{{\sqrt 3 }} + \frac{{5 \times 1}}{{12 \times 1}} \cr & \Rightarrow \frac{1}{4} - \frac{2}{3} + \frac{5}{{12}} \cr & \Rightarrow \frac{{3 - 8 + 5}}{{12}} \cr & \Rightarrow 0 \cr & {\text{ }} \cr} $$

$$\eqalign{ & \frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}} + {\sec ^2}{45^ \circ } - {\sec ^2}{0^ \circ } = x\sec {60^ \circ } \cr & \Rightarrow \frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} + {\left( {\sqrt 2 } \right)^2} - 1 = x \times 2 \cr & \Rightarrow \frac{{2 \times \frac{1}{3}}}{{1 - \frac{1}{3}}} + 2 - 1 = 2x \cr & \Rightarrow \left( {\frac{2}{3} \times \frac{3}{2}} \right) + 2 - 1 = 2x \cr & \Rightarrow 2 = x \times 2 \cr & \Rightarrow x = 1 \cr} $$

$${\text{ }}x\sin {60^ \circ }.\tan {30^ \circ } - \tan^2 {45^ \circ } = $$       $$\operatorname{cosec} {60^ \circ }.$$   $$\cot {30^ \circ } - $$   $${\sec ^2}{45^ \circ }$$
$$\eqalign{ & \Rightarrow x \times \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 3 }} - 1 = \frac{2}{{\sqrt 3 }} \times \sqrt 3 - {\left( {\sqrt 2 } \right)^2} \cr & \Rightarrow \frac{x}{2} - 1 = 2 - 2 \cr & \Rightarrow \frac{x}{2} - 1 = 0 \cr & \Rightarrow \frac{x}{2} = 1 \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & \sin \theta + \cos \theta = \sqrt 2 \sin \left( {{{90}^ \circ } - \theta } \right) \cr & \sin \theta + \cos \theta = \sqrt 2 cos\theta \cr & {\text{Divide both sides by cos}}\theta \cr & {\text{tan}}\theta + 1 = \sqrt 2 \cr & \cot \theta = \frac{1}{{\sqrt 2 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 + 1 \cr} $$

$$\eqalign{ & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & {\text{Find }}{\sin ^4}\theta = ? \cr & \frac{{\sin \theta + \cos \theta }}{{\sin \theta - \cos \theta }} = \frac{3}{1} \cr & \left( {{\text{by C & D}}} \right) \cr & \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{{3 + 1}}{{3 - 1}} \cr & \Rightarrow {\text{tan}}\theta = 2 \cr & {\text{tan}}\theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{1} \cr & \Rightarrow {\sin ^4}\theta \Rightarrow {\left( {\frac{2}{{\sqrt 5 }}} \right)^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$

$$\eqalign{ & \tan \theta + \sin \theta = m \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta + 2{\text{ tan}}\theta .\sin \theta = {m^2}\,....(i) \cr & {\text{tan}}\theta - \sin \theta = n \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta - 2{\text{ tan}}\theta .\sin \theta = {n^2}\,....(ii) \cr & {\text{Substract from (i) and (ii)}} \cr & {m^2} - {n^2} = {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta - {\text{ta}}{{\text{n}}^2}\theta - {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta \cr & {m^2} - {n^2} = 4{\text{tan}}\theta \sin \theta \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta {{\sin }^2}\theta } \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta - {{\sin }^2}\theta } \cr & = 4\sqrt {mn} \cr} $$

$${\text{ }}x{\sin ^2}{60^ \circ } - \frac{3}{2}{\text{sec}}{60^ \circ }{\text{.ta}}{{\text{n}}^2}{30^ \circ } + $$       $$\frac{4}{5}{\sin ^2}{45^ \circ }.$$   $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$   $$ = 0$$
$$ \Rightarrow {\text{ }}x{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - \frac{3}{2} \times {\text{2}} \times {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}{\text{ + }}$$       $$\frac{4}{5}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times $$   $${\left( {\sqrt 3 } \right)^2} = 0$$
$$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{3}{2} \times 2 \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{2} \times 3 = 0 \cr & \Rightarrow \frac{{3x}}{4} - 1 + \frac{6}{5} = 0 \cr & \Rightarrow \frac{{3x}}{4} = 1 - \frac{6}{5} \cr & \Rightarrow \frac{{5 - 6}}{5} \cr & \Rightarrow \frac{{ - 1}}{5} \cr & \therefore x = - \frac{1}{5} \times \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\, = - \frac{4}{{15}} \cr} $$

$$\eqalign{ & {\text{ta}}{{\text{n}}^2}{\text{A}} + {\text{co}}{{\text{t}}^2}{\text{A}} - {\sec ^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}{45^ \circ } + {\text{co}}{{\text{t}}^2}{45^ \circ } - {\sec ^2}{45^ \circ }{\text{.cose}}{{\text{c}}^2}{45^ \circ } \cr & \Rightarrow 1 + 1 - {\left( {\sqrt 2 } \right)^2}{\left( {\sqrt 2 } \right)^2} \cr & \Rightarrow - 2 \cr} $$

$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ } + 4{\text{se}}{{\text{c}}^2}{{30}^ \circ } - {\text{ta}}{{\text{n}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {\text{co}}{{\text{s}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{2}} \right)}^2} + 4{{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2} - 1}}{1} \cr & \left( {{{\sin }^2}{\text{A}} + {{\cos }^2}{\text{A}} = {\text{1}}} \right) \cr & \Rightarrow \frac{1}{4} + \frac{{4 \times 4}}{3} - 1 \cr & \Rightarrow \frac{1}{4} + \frac{{16}}{3} - 1 \cr & \Rightarrow \frac{{3 + 64 - 12}}{{12}} \cr & \Rightarrow \frac{{55}}{{12}} \cr} $$

$${\text{152}}\left( {\sin {{30}^ \circ }\, + \,2{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,3\sin {{30}^ \circ }\, + \,4{\text{co}}{{\text{s}}^2}{{45}^ \circ }\, + \,.....\, + \,17\sin {{30}^ \circ }\, + \,18{\text{co}}{{\text{s}}^2}{{45}^ \circ }} \right)$$
$$ = 152\left\{ {\frac{1}{2}\, + \,2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}\, + \,3\, \times \,\frac{1}{2}\, + \,.....\,\, + \,17 \times \frac{1}{2}\, + \,18{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right\}$$
$$\eqalign{ & = 152\left\{ {\frac{1}{2}\, + \,1\, + \,1\frac{1}{2} + \,.....\,8\frac{1}{2}\, + \,9} \right\} \cr & = {\text{This is in A}}{\text{.P}}{\text{. where}} \cr & a = \frac{1}{2},{\text{ }}d = \frac{1}{2},{\text{ }}n = 18 \cr & = 152\left\{ {\frac{{18}}{2}\left( {2\, \times \,\frac{1}{2}\, + \,\left( {18\, - \,1} \right)\frac{1}{2}} \right)} \right\} \cr & = 152\left\{ {\frac{{18}}{2}\left( {1\, + \,\frac{{17}}{2}} \right)} \right\} \cr & = 152 \times 9 \times \frac{{19}}{2} \cr & = 12996 \cr & = \sqrt {12996} \cr & = 114 \cr} $$

$$\eqalign{ & {\text{sin}}\left( {4\alpha - \beta } \right) = 1 = \sin {90^ \circ } \cr & 4\alpha - \beta = {90^ \circ }{\text{ }} \cr & {\text{cos}}\left( {2\alpha + \beta } \right) = \frac{1}{2} = \cos {60^ \circ } \cr & 2\alpha + \beta = {60^ \circ } \cr & {\text{Adding }}6\alpha = {150^ \circ } \cr & \alpha = {25^ \circ } \cr & \beta = {10^ \circ } \cr & \Rightarrow {\text{sin}}\left( {\alpha + 2\beta } \right) \cr & \Rightarrow {\text{sin}}\left( {{{25}^ \circ } + 2 \times {{10}^ \circ }} \right) \cr & \Rightarrow {\text{sin}}{45^ \circ } \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{\text{A}} - {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} - {\text{B}} = {\text{3}}{0^ \circ }} \right) \cr & {\text{cos}}\left( {{\text{A}} + {\text{B}}} \right) = \frac{1}{2}{\text{ }}\left( {{\text{A}} + {\text{B}} = {{60}^ \circ }} \right) \cr & {\text{Adding}}\,{\text{both}}\,{\text{side}} \cr & \Rightarrow \left( {{\text{A}} - {\text{B}}} \right) + \left( {{\text{A + B}}} \right) = {30^ \circ } + {60^ \circ } \cr & \Rightarrow 2{\text{A}} = {90^ \circ } \cr & \Rightarrow {\text{A}} = {45^ \circ } \cr & \because {\text{A}} - {\text{B}} = {30^ \circ } \cr & {\text{B}} = {\text{A}} - {30^ \circ } \cr & \Rightarrow {45^ \circ } - {30^ \circ } \cr & \Rightarrow {15^ \circ } \cr & \Leftrightarrow \frac{{15 \times \pi }}{{180}} = \frac{\pi }{{12}}({\text{radian}}) \cr} $$

$$\eqalign{ & x\sin {45^ \circ } = y\operatorname{cosec} {30^ \circ } \cr & \Rightarrow \frac{x}{y} = \frac{{{\text{cosec 3}}{{\text{0}}^ \circ }}}{{{\text{sin }}{{45}^ \circ }}} \cr & \Rightarrow \frac{x}{y} = \frac{2}{{\frac{1}{{\sqrt 2 }}}} \cr & \Rightarrow \frac{x}{y} = \frac{{2\sqrt 2 }}{1} \cr & \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {\left( {\frac{{2\sqrt 2 }}{1}} \right)^4} \cr & \Rightarrow \frac{{{x^4}}}{{{y^4}}} = \frac{{64}}{1} \cr & \Rightarrow \frac{{{x^4}}}{{{y^4}}} = {4^3} \cr} $$

$$\eqalign{ & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & \Rightarrow {\text{se}}{{\text{c}}^2}\alpha - 1 = 1 + 2\left( {{\text{se}}{{\text{c}}^2}\beta - 1} \right) \cr & \Rightarrow {\sec ^2}\alpha - 1 = 2{\sec ^2}\beta - 1 \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}\alpha }} = \frac{2}{{{\text{co}}{{\text{s}}^2}\beta }} \cr & \Rightarrow \sqrt 2 {\text{cos}}\alpha = {\text{cos}}\beta \cr & \therefore \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2{\text{ta}}{{\text{n}}^2}\beta \cr & {\text{Put }}\beta = {45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 1 + 2.{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr & {\text{ ta}}{{\text{n}}^2}\alpha = 3 \cr & {\text{ tan }}\alpha = \sqrt 3 \cr & \alpha = {60^ \circ } \cr & {\text{Put }}\alpha = {60^ \circ },{\text{and }}\beta = {45^ \circ } \cr & = \sqrt 2 {\text{cos}}\alpha - {\text{cos}}\beta \cr & = \sqrt 2 {\text{cos }}{60^ \circ } - {\text{cos }}{45^ \circ } \cr & = \sqrt 2 \times \frac{1}{2} - \frac{1}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0 \cr} $$