$$\eqalign{ & r\sin \theta = 1,{\text{ }}r\cos \theta = \sqrt 3 \cr & {\text{Put }}\theta = {30^ \circ } \cr & r = 2 \cr & {\text{So}},{\text{ }}{r^2}\tan \theta \cr & = {\left( 2 \right)^2} \times {\text{tan}}{30^ \circ } \cr & = 4 \times \frac{1}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

Shortcut method :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$     $$\left( {1 + {{\cot }^2}\beta } \right)$$  $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{ & \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr & {\text{Put }} \alpha = \beta = {45^ \circ } \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{45^ \circ }.{\sin ^2}{45^ \circ }.{\text{cose}}{{\text{c}}^2}{45^ \circ }.{\operatorname{sce} ^2}{45^ \circ } \cr & \Rightarrow \frac{1}{2}.\frac{1}{2}.2.2 \cr & \Rightarrow 1 \cr} $$

Alternate :
$$\left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\cos }^2}\alpha } \right) \times $$     $$\left( {1 + {{\cot }^2}\beta } \right)$$  $$\left( {1 + {{\tan }^2}\beta } \right)$$
$$\eqalign{ & \Rightarrow \left( {{{\cos }^2}\alpha } \right)\left( {si{n^2}\alpha } \right) \times \left( {{{\operatorname{cosec} }^2}\beta } \right)\left( {{{\sec }^2}\beta } \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\left( {{{90}^ \circ } - \beta } \right).{\sin ^2}\alpha .{\text{cose}}{{\text{c}}^2}\beta .{\text{se}}{{\text{c}}^2}\left( {{{90}^ \circ } - \alpha } \right) \cr & \Rightarrow {\sin ^2}\beta .{\text{cose}}{{\text{c}}^2}\beta .{\sin ^2}\alpha . {\text{cose}}{{\text{c}} ^2}\alpha \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By componendo and dividendo}} \cr & \Rightarrow \frac{{2{\text{tan}}\theta }}{{2{\text{cos}}\theta }} = \frac{3}{1} \cr & \Rightarrow \frac{{\sin \theta }}{{{\text{cos}}\theta }} \times \frac{{\sin \theta }}{{{\text{cos}}\theta }} = 3 \cr & \Rightarrow {\sin ^2}\theta = 3{\text{co}}{{\text{s}}^2}\theta \cr & \Rightarrow {\sin ^2}\theta = 3\left( {1 - {{\sin }^2}\theta } \right) \cr & \Rightarrow 4{\sin ^2}\theta = 3 \cr & \Rightarrow {\sin ^2}\theta \Rightarrow \frac{3}{4} \cr & \Rightarrow {\text{sin }}\theta = \frac{{\sqrt 3 }}{2} \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow \frac{{{\text{tan}}\theta + \cot \theta }}{{{\text{tan}}\theta - \cot \theta }} = 2 \cr & {\text{By C and D}} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{\cot \theta }} = \frac{3}{1} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = 3 \cr & \Rightarrow {\text{tan}}\theta = \sqrt 3 \cr & \theta = {60^ \circ } \cr & \Rightarrow \sin \theta \cr & \Rightarrow {\text{sin }}{60^ \circ } \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \cr} $$

$$\eqalign{ & {\sin ^2}\theta - 3\sin \theta + 2 = 0 \cr & \Rightarrow {\sin ^2}\theta - 2{\text{sin }}\theta - \sin \theta + 2 = 0 \cr & \Rightarrow \sin \theta \left( {\sin \theta - 2} \right) - 1\left( {\sin \theta - 2} \right) = 0 \cr & \Rightarrow \left( {\sin \theta - 1} \right)\left( {\sin \theta - 2} \right) = 0 \cr & \left[ {\because \sin \theta \ne 2} \right]{\text{Put value of }} \cr & \Rightarrow \sin \theta = 1 \cr & \Rightarrow {\text{sin }}\theta = \sin {90^ \circ } \cr & \Rightarrow \theta = {90^ \circ } \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }}\theta = {90^ \circ } \cr & \left[ {{\text{Take help from the options}}} \right] \cr & \Rightarrow {\sin ^2}\theta - 3\sin {\text{ }}\theta + 2 = 0 \cr & \Rightarrow {\sin ^2}{90^ \circ } - 3{\text{sin }}{90^ \circ } + 2 = 0 \cr & \Rightarrow 1 - 3 \times 1 + 2 = 0 \cr & \left[ {\sin {{90}^ \circ } = 1} \right] \cr & \Rightarrow 0 = 0\left[ {{\text{matched}}} \right] \cr & {\text{So, this is answer}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{2}}\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{1}{2} \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta = 1 + \frac{1}{2} \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta = \frac{3}{2} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{3}{4} \cr & \Rightarrow {\text{se}}{{\text{c}}^2}\theta = \frac{4}{3} \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}\theta = \frac{4}{3} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{4}{3} - 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{1}{3} \cr & \Rightarrow \tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \cot \theta = \sqrt 3 \cr & \cr & {\bf{Alternate:}} \cr & {\text{2}}\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {{\text{1}} - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{1}{2} \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta = 1 + \frac{1}{2} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{3}{2} \times \frac{1}{2} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \frac{3}{4} \cr & \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{2}\left[ {\cos {{30}^ \circ } = \frac{{\sqrt 3 }}{2}} \right] \cr & \Rightarrow \theta = {30^ \circ } \cr & {\text{Hence, }} \cr & {\text{cot}}\theta = {\text{cot3}}{0^ \circ } = \sqrt 3 \cr} $$

$$\eqalign{ & {\text{tan}}\theta + {\text{cot}}\theta = 2 \cr & {\text{Put }}\theta = {45^ \circ } \cr & 1 + 1 = 2\left( {{\text{matched}}} \right) \cr & {\text{So, }}\theta = {45^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^{100}}{45^ \circ } + {\text{co}}{{\text{t}}^{100}}{45^ \circ } \cr & \Rightarrow {1^{100}} + {1^{100}} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{sec}}\theta \left( {\frac{{1 + \sin \theta }}{{{\text{cos}}\theta }} + \frac{{{\text{cos}}\theta }}{{1 + \sin \theta }}} \right) - 2{\text{ta}}{{\text{n}}^2}\theta \cr & {\bf{Shortcut method:}} \cr & {\text{Take, }}\theta = {0^ \circ } \cr & \Rightarrow {\text{sec }}{0^ \circ }\left( {\frac{{1 + \sin {0^ \circ }}}{{{\text{cos }}{0^ \circ }}} + \frac{{{\text{cos }}{0^ \circ }}}{{1 + \sin {0^ \circ }}}} \right) - 2{\text{ta}}{{\text{n}}^2}{0^ \circ } \cr & \Rightarrow 1\left( {\frac{{1 + 0}}{1} + \frac{1}{{1 + 0}}} \right) - 0 \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{ tan }}{9^ \circ } = \frac{p}{q} \cr & \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{1 + {{\cot }^2}{{81}^ \circ }}} \cr & \Rightarrow \frac{{{\text{se}}{{\text{c}}^2}{\text{8}}{{\text{1}}^ \circ }}}{{{{\operatorname{cosec} }^2}{{81}^ \circ }}} \cr & \Rightarrow \frac{1}{{{\text{co}}{{\text{s}}^2}{{81}^ \circ }}} \times {\sin ^2}{81^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}{81^ \circ } \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\left( {{{90}^ \circ } - {9^ \circ }} \right) \cr & \Rightarrow {\text{co}}{{\text{t}}^2}{9^ \circ } \cr & \Rightarrow \frac{{{q^2}}}{{{p^2}}} \cr} $$

The value of,
cos24° + cos55° + cos125° + cos204° + cos300°
We know that, cos(180° $$ \pm $$ θ) = -cosθ
⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°)
⇒ cos24° + cos55° - cos55° - cos24° + cos60°
⇒ cos60°
⇒ $$\frac{1}{2}$$

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$

sin²1° + sin²11° + sin²21° + sin²31° + sin²41° + sin²45° + sin²49° + sin²59° + sin²69° + sin²79° + sin²89°
= (sin²1° + sin²89°) + (sin²11° + sin²79°) + (sin²21° + sin²69°) + (sin²31° + sin²59°) + (sin²41° + sin²49°) + sin²45°
= 1 + 1 + 1 + 1 + 1 + $$\frac{1}{2}$$   [sin²A + sin²B = 1. If, A + B = 90°]
= $$5\frac{1}{2}$$

$$\eqalign{ & \cos {20^ \circ } = m{\text{ }} \cr & \cos {70^ \circ } = n \cr & {\text{So,}} \cr & \Leftrightarrow {m^2} + {n^2} = {\text{co}}{{\text{s}}^2}{20^ \circ } + {\text{co}}{{\text{s}}^2}{70^ \circ } \cr & \left[ {{\text{If co}}{{\text{s}}^2}{\text{A + co}}{{\text{s}}^2}{\text{B}} = {\text{1}}} \right] \cr & ({\text{If, A}} + {\text{B}} = {90^ \circ }) \cr & \Leftrightarrow 1 \cr} $$

$$\eqalign{ & \Rightarrow x\tan {60^ \circ } + \cos {45^ \circ } = \sec {45^ \circ } \cr & \Rightarrow x.\sqrt 3 + \frac{1}{{\sqrt 2 }} = \sqrt 2 \cr & \Rightarrow x\sqrt 6 + 1 = 2 \cr & \Rightarrow x = \frac{1}{{\sqrt 6 }} \cr }$$
Squaring both sides and addition 1 both sides
$$\eqalign{ & \Rightarrow {x^2} + 1 = {\left( {\frac{1}{{\sqrt 6 }}} \right)^2} + 1 \cr & \Rightarrow {x^2} + 1 = \frac{1}{6} + 1 \cr & \Rightarrow {x^2} + 1 = \frac{7}{6} \cr} $$

sin(2x - 20) = cos(2y + 20)
⇒ sin(2x - 20) = sin(90 - (2y + 20))
i.e. 2x - 20 = 90 - 2y - 20
⇒ 2x + 2y = 90
⇒ x + y = 45°
∴ tan(x + y) = tan 45° = 1

$${\text{ tan 1}}{5^ \circ }{\text{cot 7}}{5^ \circ } + {\text{tan 7}}{5^ \circ }{\text{cot 1}}{5^ \circ }$$
  $$ = {\text{ tan 1}}{5^ \circ }{\text{cot }}\left( {{{90}^ \circ } - {{15}^ \circ }} \right) + $$      $${\text{tan}}{\left( {{{90}^ \circ } - 15} \right)^ \circ }$$    $${\text{cot1}}{5^ \circ }$$
$$\eqalign{ & = {\text{ ta}}{{\text{n}}^2}{\text{1}}{5^ \circ } + {\text{co}}{{\text{t}}^2}{\text{1}}{5^ \circ } \cr & = {\text{ta}}{{\text{n}}^2}{15^ \circ } + {\text{co}}{{\text{t}}^2}{15^ \circ }\,.....(i) \cr & \left[ {{\bf{Formula}}} \right] \cr & \cot \left( {{{90}^ \circ } - \theta } \right) = \tan \theta \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta \cr & {\text{Put value of tan1}}{5^ \circ } \cr & \cot {15^ \circ } = \frac{1}{{{\text{tan1}}{5^ \circ }}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}} \cr & \cot {15^ \circ } = 2 + \sqrt 3 \cr & {\text{Now put value in equation (i)}} \cr & {\text{ tan 1}}{5^ \circ } + {\text{cot 1}}{5^ \circ } \cr & = {\left( {2 - \sqrt 3 } \right)^2} + {\left( {2 + \sqrt 3 } \right)^2} \cr & = 4 + 3 - 4\sqrt 3 + 4 + 3 + 4\sqrt 3 \cr & = 14 \cr} $$

$$\eqalign{ & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2cot}}{{61}^ \circ }.\cot {{79}^ \circ }}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2}}\left[ {{\text{cot}}\left( {{{90}^ \circ } - {{29}^ \circ }} \right).\cot \left( {{{90}^ \circ } - {{11}^ \circ }} \right)} \right]}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{tan1}}{{\text{1}}^ \circ }{\text{.tan2}}{{\text{9}}^ \circ }}}{{{\text{2tan1}}{{\text{1}}^ \circ }.tan{{29}^ \circ }}} \cr & \Leftrightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{1}{2} \cr & \Leftrightarrow 2{\text{A}} = {\text{B}} \cr} $$

$$\eqalign{ & \sin {\text{ }}\alpha + \cos \beta = 2 \cr & {\bf{Shortcut\,\, method:}} \cr & {\text{Put, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Leftrightarrow {\text{sin }}{90^ \circ } + {\text{cos }}{0^ \circ } = 2 \cr & \Leftrightarrow 1 + 1 = 2 \cr & \Leftrightarrow 2 = 2\left[ {{\text{Matched}}} \right] \cr & {\text{So, }}\alpha = {90^ \circ },\beta = {0^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2\alpha + \beta }}{3}} \right)^ \circ }{\text{ }} \cr & \Rightarrow {\text{sin }}{\left( {\frac{{2 \times {{90}^ \circ } + {0^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{\left( {\frac{{{{180}^ \circ }}}{3}} \right)^ \circ } \cr & \Rightarrow {\text{sin }}{60^ \circ } = \cos {30^ \circ } = \frac{{\sqrt 3 }}{2} \cr & {\text{Take cos}}\frac{\alpha }{3} = \cos \frac{{{{90}^ \circ }}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{cos 3}}{0^ \circ } \cr & {\text{So, this is answer }}{\text{.}} \cr} $$

$$\eqalign{ & 2\sin \left( {\frac{{\pi x}}{2}} \right) = {x^2} + \frac{1}{{{x^2}}} \cr & {\text{Let }}x = 1 \cr & \Rightarrow 2\sin {90^ \circ } = {1^2} + \frac{1}{{{1^2}}} \cr & \Rightarrow 2 \times 1 = 1 + 1 \cr & 2 = 2\left( {{\text{Matched}}} \right) \cr & {\text{So, }}x = 1 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) \cr & \Rightarrow 1 - \frac{1}{1} \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {\text{secA}} + {\text{tanA}} = a \cr & {\text{we know that}} \cr & \Rightarrow {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1 \cr & \Rightarrow \left( {{\text{secA}} - {\text{tanA}}} \right)\left( {{\text{secA}} + {\text{tanA}}} \right) = 1 \cr & \Rightarrow {\text{secA}} - {\text{tanA}} = \frac{1}{a} \cr & \Rightarrow {\text{secA}} + {\text{tanA}} = a \cr & \Rightarrow {\text{2secA}} = a + \frac{1}{a} \cr & \Rightarrow {\text{2secA}} = \frac{{{a^2} + 1}}{a} \cr & \Rightarrow \sec \theta = \frac{{{a^2} + 1}}{{2a}} \cr & {\text{So, }}\cos \theta = \frac{{2a}}{{{a^2} + 1}} \cr} $$

$$\eqalign{ & {\text{sin P}} + {\text{cosec P}} = 2 \cr & {\text{For P}} = {90^ \circ } \cr & \Rightarrow {\text{sin }}{90^ \circ } + {\text{cosec }}{90^ \circ } = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & \Rightarrow 2 = 2\left( {{\text{satisfy}}} \right) \cr & {\text{So, }}{\sin ^7}{\text{P}} + {\text{cose}}{{\text{c}}^7}{\text{P}} \cr & \Rightarrow {\sin ^7}{90^ \circ } + {\text{cose}}{{\text{c}}^7}{90^ \circ } \cr & \Rightarrow {1^7} + {1^7} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{tan}}\left( {2\theta + {{45}^ \circ }} \right) = \cot 3\theta \cr & \left[ {{\text{If tan A}} = {\text{cot B}}} \right] \cr & ({\text{then, A}} + {\text{B}} = {90^ \circ }) \cr & \Rightarrow \left( {2\theta + {{45}^ \circ }} \right) + 3\theta = {90^ \circ } \cr & \Rightarrow 5\theta + {45^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = \frac{{45}}{5} \cr & \Rightarrow \theta = {9^ \circ } \cr} $$

$$\eqalign{ & {\text{sin}}\left( {{{90}^ \circ } - \theta } \right) + {\text{cos}}\theta = \sqrt 2 {\text{cos}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{cos}}\theta + {\text{cos}}\theta = \sqrt 2 \sin \theta \cr & \Rightarrow \frac{{2\cos \theta }}{{\sin \theta }} = \sqrt 2 \cr & \Rightarrow \cot \theta = \frac{{1 \to {\text{B}}}}{{\sqrt 2 \to {\text{P}}}} \cr & {\text{So, H}} \to \text{alignment} \cr & \therefore {\text{cosec}}\theta = \frac{{\text{H}}}{{\text{P}}} = \frac{{\sqrt 3 }}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{3}{2}} \cr} $$

$$\eqalign{ & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\bf{Shortcut \,\,method:}} \cr & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow {\text{sin }}{60^ \circ } - \cos {60^ \circ } = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} - \frac{1}{2} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 - 1}}{2} = \frac{{\sqrt 3 - 1}}{2}({\text{Matched}}) \cr & Hence, \cr & {\text{sin A}}.cos{\text{A}} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \times \frac{1}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\text{Squaring both side,}} \cr & \Rightarrow {\text{ si}}{{\text{n}}^2}{\text{ A + }}{\cos ^2}{\text{A}} - {\text{2}}{\text{.sin A}}.cos{\text{A}} = {\left( {\frac{{\sqrt 3 - 1}}{2}} \right)^2} \cr & \Rightarrow 1 - 2{\text{sin A}}.cos{\text{A = }}\frac{{3 + 1 - 2\sqrt 3 }}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = 1 - 2\frac{{\left( {2 - \sqrt 3 } \right)}}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = \frac{{2 - 2 + \sqrt 3 }}{2} \cr & \Rightarrow {\text{sin A}}.cos{\text{A}} = \frac{{\sqrt 3 }}{4} \cr} $$

$${\text{ 2}}\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) - {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right){\text{ + 1}}$$
$$ \Rightarrow {\text{2}}\left( {1 - 3{{\sin }^2}\theta {\text{co}}{{\text{s}}^2}\theta } \right) - $$     $${\text{3}}\left( {1 - 2{{\sin }^2}\theta c{\text{o}}{{\text{s}}^2}\theta } \right){\text{ + }}$$     $${\text{1}}$$
$$\eqalign{ & \Rightarrow {\text{ 2}} - 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta - 3 + 6{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta {\text{ + 1}} \cr & \Rightarrow 2 - 3 + 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {a^2}{\sec ^2}x - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2}\left( {1 + ta{n^2}x} \right) - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2} + {a^2}{\tan ^2}x - {b^2}{\tan ^2}x = {c^2} \cr & \Rightarrow {a^2} + ta{n^2}x\left( {{a^2} - {b^2}} \right) = {c^2} \cr & \Rightarrow {a^2} - {c^2} = {\tan ^2}x\left( {{b^2} - {a^2}} \right) \cr & \Rightarrow \frac{{{a^2} - {c^2}}}{{{b^2} + {a^2}}} = {\text{ta}}{{\text{n}}^2}x \cr & \Rightarrow {\sec ^2}x - {\tan ^2}x = 1 \cr & \Rightarrow {\sec ^2}x = {\tan ^2}x + 1 \cr & \Rightarrow 1 + \frac{{{a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & \Rightarrow \frac{{{b^2} - {a^2} + {a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & \Rightarrow \frac{{{b^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & {\sec ^2}x + {\tan ^2}x \cr & = \frac{{{b^2} - {c^2}}}{{{b^2} - {a^2}}} + \frac{{{a^2} - {c^2}}}{{{b^2} - {a^2}}} \cr & = \frac{{{b^2} + {a^2} - 2{c^2}}}{{{b^2} - {a^2}}} \cr} $$

$$\eqalign{ & {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta = 1\,......({\text{i}}) \cr & \because {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta \left( {{{\sec }^2}\theta } \right) = 1 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = \frac{1}{{{{\sec }^2}\theta }} \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = {\text{co}}{{\text{s}}^2}\theta \cr & \because {\text{ co}}{{\text{s}}^4}\theta + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{co}}{{\text{s}}^2}\theta } \right)^2} + {\text{co}}{{\text{s}}^2}\theta \cr & = {\left( {{\text{ta}}{{\text{n}}^2}\theta } \right)^2} + {\text{ta}}{{\text{n}}^2}\theta \cr & = {\text{ta}}{{\text{n}}^4}\theta + {\text{ta}}{{\text{n}}^2}\theta \cr & = 1{\text{ from equation }}\left( {\text{i}} \right) \cr} $$

$$\eqalign{ & {\text{ tan}}\left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) = \sqrt 3 \cr & \Rightarrow {\text{tan}}\left( {{{90}^ \circ } - \frac{\theta }{2}} \right) = \sqrt 3 \,\,\left[ {\pi = {{180}^ \circ }} \right] \cr & \Rightarrow \cot \frac{\theta }{2} = \sqrt 3 \cr & \Rightarrow \cot \frac{\theta }{2} = \cot {30^ \circ } \cr & \Rightarrow \frac{\theta }{2} = {30^ \circ } \cr & \Rightarrow \theta = {60^ \circ } \cr & \Rightarrow \cos {60^ \circ } = \frac{1}{2} \cr} $$

$$\eqalign{ & \sqrt {\frac{{1 + {\text{sin }}\theta }}{{1 - {\text{sin }}\theta }}} {\text{ + }}\sqrt {\frac{{1 - {\text{sin }}\theta }}{{1 + {\text{sin }}\theta }}} \cr & = \frac{{{{\left( {\sqrt {1 + {\text{sin }}\theta } } \right)}^2} + {{\left( {\sqrt {1 - {\text{sin }}\theta } } \right)}^2}}}{{\sqrt {1 - {{\sin }^2}\theta } }} \cr & = \frac{{1 + {\text{sin }}\theta + {\text{1}} - {\text{sin }}\theta }}{{{\text{cos}}\theta }} \cr & = \frac{2}{{{\text{cos}}\theta }} \cr & = 2{\text{ sec}}\theta \cr} $$

Alternate shortcut method :
$$\eqalign{ & {\text{Put }}\theta = {30^ \circ } \cr & \sqrt {\frac{{1 + {\text{sin }}30^ \circ }}{{1 - {\text{sin }}30^ \circ }}} {\text{ + }}\sqrt {\frac{{1 - {\text{sin }}30^ \circ }}{{1 + {\text{sin }}30^ \circ }}} \cr & \Rightarrow \sqrt {\frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}}} + \sqrt {\frac{{1 - \frac{1}{2}}}{{1 + \frac{1}{2}}}} \cr & \Rightarrow \sqrt {\frac{3}{1}} + \sqrt {\frac{1}{3}} \cr & \Rightarrow \frac{4}{{\sqrt 3 }} \cr} $$
Now check with option by puting θ = 30°
$$\eqalign{ & {\text{ = 2 sec 3}}{0^ \circ } \cr & = \frac{{2 \times 2}}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {\left( {r\cos \theta - \sqrt 3 } \right)^2} + {\left( {r\sin \theta - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {r\cos \theta - \sqrt 3 } \right)^2} = 0,{\text{ }}{\left( {r\sin \theta - 1} \right)^2} = 0 \cr & \Rightarrow r{\text{ cos}}\theta = \sqrt 3 ........(i) \cr & \Rightarrow r\sin \theta = 1.........(ii) \cr & {\text{Squaring and adding equation (i) and (ii)}} \cr & \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 3 + 1 \cr & \Rightarrow {r^2}\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {r^2} = 4 \cr & \Rightarrow r = 2 \cr & {\text{tan}}\theta = \frac{{r\sin \theta }}{{r{\text{ cos}}\theta }} = \frac{1}{{\sqrt 3 }}{\text{ and }}r{\text{ cos}}\theta = \sqrt 3 \cr & \cos \theta = \frac{{\sqrt 3 }}{r}, \cr & \sec \theta = \frac{r}{{\sqrt 3 }} \cr & \frac{{r\tan \theta + \sec \theta }}{{r\sec \theta + \tan \theta }} \cr & = \frac{{\frac{r}{{\sqrt 3 }} + \frac{r}{{\sqrt 3 }}}}{{\frac{{{r^2}}}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }}}} \cr & = \frac{{r\left( {\frac{2}{{\sqrt 3 }}} \right)}}{{\frac{{{r^2} + 1}}{{\sqrt 3 }}}} \cr & = \frac{{2r}}{{{r^2} + 1}} \cr & = \frac{{2 \times 2}}{{{2^2} + 1}} \cr & = \frac{4}{5} \cr & \cr & {\bf{Alternate:}} \cr & r = 2 \cr & {\text{tan}}\theta = \frac{{r\sin \theta }}{{r{\text{ cos}}\theta }} = \frac{1}{{\sqrt 3 }} \cr & \theta = {30^ \circ } \cr & = \frac{{2\tan {{30}^ \circ } + \sec {{30}^ \circ }}}{{2\sec {{30}^ \circ } + \tan {{30}^ \circ }}} \cr & = \frac{{2 \times \frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }}}}{{2 \times \frac{2}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }}}} \cr & = \frac{4}{5} \cr} $$

$$\eqalign{ & x = {\text{cosec}}\theta - \sin \theta {\text{ }} \cr & y = \sec \theta - \cos \theta \cr & {\text{Put }}\theta = {45^ \circ } \cr & x = \sqrt 2 - \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & y = \sqrt 2 - \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & {\text{by options (B) }}{x^2}{y^2}\left( {{x^2} + {y^2} + 3} \right) \cr & = {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & \left[ {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + 3} \right] \cr & = \frac{1}{2} \times \frac{1}{2}\left( {\frac{1}{2} + \frac{1}{2} + 3} \right) \cr & = \frac{1}{4}\left( {1 + 3} \right) \cr & = 1{\text{ }}\left( {{\text{satisfy}}} \right) \cr} $$