Answer: Option A. -> x2 + y2 = 20
$$\eqalign{
& xsin\theta + y\cos \theta = 4 \cr
& \underline {x\cos \theta - y\sin \theta = 2} {\text{ }}\,\,\,\,\left[ {{\text{Formula}}} \right] \cr
& \left( {{x^2} + {y^2}} \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = {4^2} + {2^2} \cr
& \Rightarrow {x^2} + {y^2} = {a^2} + {b^2} \cr
& \Rightarrow \left( {{x^2} + {y^2}} \right)\left( 1 \right) = 16 + 4 \cr
& \Rightarrow {x^2} + {y^2} = 20 \cr} $$

Answer: Option C. -> $$\frac{2}{3}$$
$$\eqalign{
& {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{2}{3} \cr
& \Rightarrow \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) = \frac{2}{3} \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = \frac{2}{3} \cr
& \Rightarrow 1 - {\sin ^2}\theta - {\sin ^2}\theta = \frac{2}{3} \cr
& \Rightarrow 1 - 2{\sin ^2}\theta = \frac{2}{3} \cr} $$

Answer: Option A. -> 2 cosec A
$$\eqalign{
& \frac{{{\text{sin A}}}}{{1 + \cos {\text{ A}}}}{\text{ + }}\frac{{{\text{sin A}}}}{{1 - \cos {\text{ A}}}} \cr
& \Rightarrow \frac{{{\text{sin A}}\left( {1 - \cos {\text{ A}}} \right) + {\text{sin A}}\left( {1 + \cos {\text{ A}}} \right)}}{{\left( {1 + \cos {\text{ A}}} \right)\left( {1 - \cos {\text{ A}}} \right)}} \cr
& \Rightarrow \frac{{{\text{sin A}} - {\text{sin A}}{\text{.cosA}} + {\text{sin A}} + {\text{sin A}}{\text{.cosA}}}}{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr
& \Rightarrow \frac{{2{\text{sin A}}}}{{{{\sin }^2}{\text{A}}}} \cr
& \Rightarrow 2{\text{ cosec A}} \cr} $$

Answer: Option C. -> 2
  $$\left[ {\frac{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.si}}{{\text{n}}^2}{\text{A}}\left( {{\text{sin A}} + {\text{cos A}}} \right)}}{{\left( {{\text{sin A}} - {\text{cos A}}} \right)}} + \frac{{{\text{si}}{{\text{n}}^2}{\text{A}}{\text{.co}}{{\text{s}}^2}{\text{A}}\left( {{\text{sin A}} - {\text{cos A}}} \right)}}{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}}} \right]$$           $$\left[ {\frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} - \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}}} \right]$$
  $$ \Rightarrow \left[ {\frac{{{{\left( {{\text{sin A}} + {\text{cos A}}} \right)}^2} + {{\left( {{\text{sin A}} - {\text{cos A}}} \right)}^2}}}{{\left( {{\text{sin A}} - {\text{cos A}}} \right)\left( {{\text{sin A}} + {\text{cos A}}} \right)}}} \right]$$       $$\left( {{\text{si}}{{\text{n}}^2}{\text{ A}} - {\text{co}}{{\text{s}}^2}{\text{ A}}} \right)$$
$$\eqalign{
& \Rightarrow 2\left( {{\text{si}}{{\text{n}}^2}{\text{ A}} + {\text{co}}{{\text{s}}^2}{\text{ A}}} \right) \cr
& \Rightarrow 2 \cr} $$

Answer: Option D. -> 2
$$\eqalign{
& r\sin \theta = 1 \cr
& r\cos \theta = \sqrt 3 \cr
& \Rightarrow \frac{{{\text{ }}rsin\theta }}{{{\text{ }}r\cos \theta }} = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow {\text{tan}}\theta = \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \sqrt 3 {\text{tan}}\theta = 1 \cr
& \left( {{\text{Add 1 both sides}}} \right) \cr
& \Rightarrow \sqrt 3 {\text{tan}}\theta + 1 = 1 + 1 \cr
& \Rightarrow \sqrt 3 {\text{tan}}\theta + 1 = 2 \cr} $$

Answer: Option C. -> 0
$$\eqalign{
& {\sin ^{12}} + 3{\sin ^{10}}x + 3{\sin ^8}x + {\sin ^6}x - 1 \cr
& \Rightarrow {\left( {{{\sin }^4}x + {{\sin }^2}x} \right)^3} - 1 \cr
& \Rightarrow {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)^3} - 1 \cr
& \left[ {\cos x + {{\cos }^2}x = 1} \right] \cr
& (\cos x = 1 - {\cos ^2}x = {\sin ^2}x) \cr
& \Rightarrow 1 - 1 \cr
& \Rightarrow 0 \cr} $$

Answer: Option C. -> $$\frac{3}{5}$$
$$\eqalign{
& {\text{According to the question,}} \cr
& {\text{cosec A}} + {\text{cot A}} = 3 \cr
& {\text{cosec A}} - {\text{cot A}} = \frac{1}{3} \cr
& \overline {2{\text{cosec A}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{3}} \cr
& {\text{cosec A}} = \frac{{10}}{6} \cr
& {\text{sin A}} = \frac{6}{{10}} \cr
& {\text{sin A}} = \frac{3}{5} \cr} $$

Answer: Option B. -> 75°, 60°
$$\eqalign{
& \angle {\text{A}} + \angle {\text{B}} = {135^ \circ }\,.....(i) \cr
& \angle {\text{A}} - \angle {\text{B}} = \frac{\pi }{{12}} = {15^ \circ }\,.....(ii) \cr
& Adding{\text{ both equations}} \cr
& \Rightarrow 2\angle {\text{A}} = {150^ \circ } \cr
& \Rightarrow \angle {\text{A}} = {75^ \circ } \cr
& \therefore \angle {\text{B}} = {60^ \circ } \cr} $$

Answer: Option D. -> $$\sqrt 2 $$ sinθ
$$\eqalign{
& {\text{cos}}\theta + \sin \theta = \sqrt 2 \cos \theta \cr
& {\text{Squaring both sides}} \cr
& {\text{co}}{{\text{s}}^2}\theta + {\sin ^2}\theta + 2\cos \theta .\sin \theta = 2{\text{co}}{{\text{s}}^2}\theta \cr
& \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2{\text{cos}}\theta .{\text{sin}}\theta \cr
& \Rightarrow {\text{co}}{{\text{s}}^2}\theta - {\sin ^2}\theta = 2\sin \theta .{\text{cos}}\theta \cr
& \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr
& \Rightarrow \left( {\cos \theta - \sin \theta } \right)\left( {\sqrt 2 \cos \theta } \right) = 2\sin \theta .{\text{cos}}\theta \cr
& \Rightarrow \cos \theta - \sin \theta = \frac{{2\sin \theta .\cos \theta }}{{\sqrt 2 \cos \theta }} \cr
& \Rightarrow \sqrt 2 \sin \theta \cr
& \cr
& {\bf{Alternate:}} \cr
& {\text{Let }}\sqrt 2 \cos \theta = \alpha \cr
& \therefore \cos \theta \pm \sin \theta = a \cr
& \cos \theta \pm \sin \theta = \sqrt {2 - {a^2}} \cr
& = \sqrt {2 - {a^2}} \cr
& = \sqrt {2 - 2{\text{co}}{{\text{s}}^2}\theta } \cr
& = \sqrt {2\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr
& = \sqrt {2{{\sin }^2}\theta } \cr
& = \sqrt 2 \sin \theta \cr} $$

Answer: Option A. -> $$\frac{1}{2}$$
$$\eqalign{
& {\text{3}}\left( {{{\sec }^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right) = 5 \cr
& {\sec ^2}\theta + {\text{ta}}{{\text{n}}^2}\theta = \frac{5}{3}\,.....(i) \cr
& {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1\,......(ii) \cr
& {\text{Adding equation (i) and (ii)}} \cr
& {\text{2}}{\sec ^2}\theta = \frac{8}{3} \cr
& sec\theta = \frac{2}{{\sqrt 3 }} \cr
& \therefore \theta = {30^ \circ } \cr
& \cos 2\theta = \cos 2\left( {{{30}^ \circ }} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos {60^ \circ } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$