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

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

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

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

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

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

Answer: Option C. -> 0
$$\eqalign{
& {\text{cos}}\theta + \sec \theta = \sqrt 3 \cr
& {\text{Cubing both sides}} \cr
& {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3{\text{cos}}\theta \sec \theta \left( {{\text{cos}}\theta + \sec \theta } \right) = 3\sqrt 3 \cr
& {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta + 3\sqrt 3 = 3\sqrt 3 \cr
& {\text{co}}{{\text{s}}^3}\theta + {\sec ^3}\theta = 0 \cr} $$

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

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

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