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

$$\sin \frac{{\pi x}}{2} = {x^2} - 2x + 2$$
Put value of x from options
$$\eqalign{ & x = 1 \cr & \sin \frac{\pi }{2} \times 1 = {1^2} - 2 \times 1 + 2 \cr & \Rightarrow \sin {90^ \circ } = 1 - 2 + 2 \cr & \Rightarrow 1 = 1 \cr} $$

Angles of triangle are in ration 2 : 7 : 11
According to the question,
⇒ 2x + 7x + 11x = 180
⇒ 20x = 180
⇒ x = 9
Angles of triangle is 18°, 63°, 99°

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\cos ^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {1 - {{\sin }^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put}}\theta = {30^ \circ } \cr & {\text{7}} \times \sin^2 {30^ \circ } + 3{\cos ^2}{30^ \circ } = 4 \cr & \Rightarrow 7 \times \frac{1}{4} + 3 \times \frac{3}{4} = 4 \cr & \Rightarrow \frac{7}{4} + \frac{9}{4} = 4 \cr & \Rightarrow \frac{{16}}{4} = 4 \cr & \Rightarrow 4 = 4\left( {{\text{Satisfied}}} \right) \cr & \therefore \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & \frac{1}{{1 + {{\cot }^2}\theta }} + \frac{3}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow \frac{1}{{{{\operatorname{cosec} }^2}\theta }} + \frac{3}{{{{\sec }^2}\theta }} + 2{\sin ^2}\theta \cr & \Rightarrow {\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta \cr & \Rightarrow 3\left( {{{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow 3\left( 1 \right) \cr & \Rightarrow 3 \cr} $$

cos1° cos2° cos3° ............. cos177° cos178° cos179°
= cos90°
= 0[0 will make whole series 0]
= 0

(x + 5)° + (2x - 3)° + (3x + 4)° = 180°
(Sum of all angles in triangle is 180°)
⇒ 6x + 6° = 180°
⇒ (x + 1) = 30°
⇒ x = 29°

$$\eqalign{ & \frac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr & = \frac{{cot{{33}^ \circ } + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr & = \frac{{\frac{1}{{\tan {{33}^ \circ }}} + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \frac{1}{{\tan {{53}^ \circ }}}}} \cr & = \frac{{1 + \tan {{53}^ \circ }.\tan {{33}^ \circ }}}{{\tan {{33}^ \circ }.\tan {{53}^ \circ } + 1}} \times \frac{{\tan {{53}^ \circ }}}{{\tan {{33}^ \circ }}} \cr & = \tan {53^ \circ }.cot{33^ \circ } \cr & = \cot {37^ \circ }.\tan {57^ \circ } \cr} $$

$$\eqalign{ & {\text{If }}\tan \theta = 1 \cr & {\text{It means }}\theta = {45^ \circ } \cr & = \frac{{8\sin \theta + 5\cos \theta }}{{{{\sin }^3}\theta - 2{{\cos }^3}\theta + 7\cos \theta }} \cr & = \frac{{8\sin {{45}^ \circ } + 5\cos {{45}^ \circ }}}{{{{\sin }^3}{{45}^ \circ } - 2{{\cos }^3}{{45}^ \circ } + 7\cos {{45}^ \circ }}} \cr & = \frac{{8 \times \frac{1}{{\sqrt 2 }} + 5 \times \frac{1}{{\sqrt 2 }}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} - 2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3} + 7\left( {\frac{1}{{\sqrt 2 }}} \right)}} \cr & = 2 \cr} $$

Let x = 2sin²θ + 3cos²θ
⇒ x = 2sin²θ + 2cos²θ + cos²θ
⇒ x = 2(sin²θ + cos²θ) + cos²θ
⇒ x = 2 + cos²θ     [since sin²θ + cos²θ = 1]
Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2

Alternative Solution:
2sin²θ + 3cos²θ
Minimum value is 2,
[If x sin²θ + y cos²θ, If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]

$$\eqalign{ & \left( {2\sin \theta + 3\cos \theta } \right) \cr & {\text{Maximum value of}} \cr & a\sin \theta + b\cos \theta \cr & = \sqrt {{a^2} + {b^2}} \cr & = \sqrt {{2^2} + {3^2}} \cr & = \sqrt {4 + 9} \cr & = \sqrt {13} \cr} $$

$$\eqalign{ & {\cos ^2}\theta = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & {\text{Max value of }}{\cos ^2}\theta = 1 \cr & \Rightarrow 1 = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & \Rightarrow 4xy = {\left( {x + y} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow 0 = {x^2} + {y^2} - 2xy \cr & \Rightarrow 0 = {\left( {x - y} \right)^2} \cr & \Rightarrow 0 = x - y \cr & \Rightarrow x = y \cr} $$

$$\eqalign{ & {\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & {\text{Squaring both sides}} \cr & {\sin ^4}\theta + {\cos ^4}\theta \cr & = 1 - 2{\sin ^2}\theta . {\cos ^2}\theta \cr & {\text{Put}}\,\theta = {90^ \circ } \cr & = 1 - 2{\sin ^2}{90^ \circ } \times {\cos ^2}{90^ \circ } \cr & = 1 - 0 \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow \cos \theta > {\cos ^2}\theta \cr & \Rightarrow \cos {60^ \circ } > {\cos ^2}{60^ \circ } \cr & \Rightarrow \frac{1}{2} > \frac{1}{4} \cr & \cos \theta > {\cos ^2}\theta \cr} $$

= (tan²α + 1 )sin²β
= (tan²45° + 1 )sin²45°
$$ = (1 + 1) {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$$
$$ = 2 \times \frac{1}{2} = 1$$

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

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

$$\eqalign{ & sec\theta + {\text{tan}}\theta = 2 + \sqrt 5 \,.....(i) \cr & {\text{se}}{{\text{c}}^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & \left( {sec\theta - {\text{tan}}\theta } \right)\left( {sec\theta + {\text{tan}}\theta } \right) = 1 \cr & \left( {sec\theta - {\text{tan}}\theta } \right) = \frac{1}{{2 + \sqrt 5 }} = \frac{1}{{\sqrt 5 + 2}} = \sqrt 5 - 2\,\,.....(ii) \cr & {\text{Adding equation }}{\text{ (i) and (ii)}} \cr & {\text{2}}sec\theta = 2 + \sqrt 5 + \sqrt 5 - 2 \cr & \Rightarrow 2sec\theta = 2\sqrt 5 \cr & \Rightarrow sec\theta = \sqrt 5 \cr & \Rightarrow {\text{cos}}\theta = \frac{1}{{\sqrt 5 }} \cr & \Rightarrow {\sin ^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = 1 - {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} \cr & \Rightarrow {\sin ^2}\theta = \frac{4}{5} \cr & \Rightarrow \sin \theta = \frac{2}{{\sqrt 5 }} \cr & \therefore \sin \theta + {\text{cos}}\theta = \frac{2}{{\sqrt 5 }} + \frac{1}{{\sqrt 5 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{\sqrt 5 }} \cr} $$

$$\eqalign{ & sec\theta = x + \frac{1}{{4x}} \cr & {\text{tan}}\theta = \sqrt {{\text{sec}}\theta - 1} \cr & {\text{tan}}\theta = \sqrt {{{\left[ {\frac{{4{x^2} + 1}}{{4x}}} \right]}^2} - 1} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} + 1} \right)}^2} - {{\left( {4x} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 + 8{x^2} - 16{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{16{x^4} + 1 - 8{x^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \sqrt {\frac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{{\left( {4x} \right)}^2}}}} \cr & {\text{tan}}\theta = \frac{{\left( {4{x^2} - 1} \right)}}{{4x}} \cr & \therefore sec\theta + {\text{tan}}\theta \cr & = \frac{{4{x^2} + 1}}{{4x}} + \frac{{4{x^2} - 1}}{{4x}} \cr & = \frac{{4{x^2} + 1 + 4{x^2} - 1}}{{4x}} \cr & = \frac{{8{x^2}}}{{4x}} \cr & = 2x \cr & \cr & {\bf{Alternate:}} \cr & sec\theta = x + \frac{1}{{4x}} \cr & {\text{Put }}x = 1 \cr & sec\theta = 1 + \frac{1}{4} = \frac{5}{4} = \frac{{\text{H}}}{{\text{B}}} \cr & \tan \theta = \frac{{\text{P}}}{{\text{B}}} = \frac{3}{4} \cr & {\text{Now, }} \cr & sec\theta + {\text{tan}}\theta \cr & = \frac{5}{4} + \frac{3}{4} \cr & = \frac{{5 + 3}}{4} \cr & = \frac{8}{4} \cr & = 2 \times 1 \cr & = 2x\left( {x = 1} \right) \cr} $$

$$\eqalign{ & {\text{2cos}}\theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & {\text{When,}} \cr & ax \mp by = m \cr & {\text{then, }}bx \mp ay = \sqrt {{a^2} + {b^2} - {m^2}} \cr & 2\cos \theta - \sin \theta = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow \cos \theta + 2\sin \theta = \sqrt {4 + 1 - \frac{1}{2}} \cr & \Rightarrow \cos \theta + 2\sin \theta = \frac{3}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & {\sec ^2}\theta + {\text{ta}}{{\text{n}}^2}\theta = 7 \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta = 7 \cr & \Rightarrow 2{\text{ta}}{{\text{n}}^2}\theta = 6 \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta = 3 \cr & \Rightarrow {\text{tan}}\theta = \sqrt 3 \cr & \Rightarrow \theta = {60^ \circ } \cr & \cr & {\bf{Alternate:}} \cr & {\text{Take help from option }} \cr & {\text{put }}\theta {\text{ = }}{60^ \circ } \cr & {\text{se}}{{\text{c}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{60^ \circ } = 7 \cr & {\left( 2 \right)^2}{\text{ + }}{\left( {\sqrt 3 } \right)^2} = 7 \cr & 7 = 7{\text{ }}\left( {{\text{matched }}} \right) \cr & {\text{So, }}\theta = {60^ \circ } \cr} $$

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

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

(sec x sec y + tan x tan y)² - (sec x tan y + tan x sec y)²
= sec²x. sec²y + tan²x. tan²y + 2sec x. sec y. tan x. tan y - sec²x. tan²y - tan²x. sec²y + 2sec x. tan y. tan x sec y
= sec²x [sec²y - tan²y] - tan²x [sec²y - tan²y]
= (sec²x - tan²x) (sec²y - tan²y)
= 1 × 1
= 1

$$\eqalign{ & {\text{According to the question, }} \cr & {\text{A}} = {\sin ^2}\theta + {\text{co}}{{\text{s}}^4}\theta \cr & {\text{Put }}\theta = {90^ \circ }{\text{ for maximum value of A}} \cr & {\text{A}} = {\sin ^2}{90^ \circ } + {\text{co}}{{\text{s}}^4}{90^ \circ } \cr & {\text{A}} = 1 + 0 \cr & {\text{A}} = 1 \cr & {\text{Put }}\theta = {45^ \circ }{\text{ for minimum value of A}} \cr & {\text{A}} = {\sin ^2}{45^ \circ } + {\text{co}}{{\text{s}}^4}{45^ \circ } \cr & {\text{A}} = \frac{1}{2} + \frac{1}{4} \cr & {\text{A}} = \frac{3}{4} \cr & \therefore {\text{A lies in }}\frac{3}{4} \leqslant {\text{A}} \leqslant {\text{1}} \cr} $$

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

$$\eqalign{ & \sin \left( {\theta + {{30}^ \circ }} \right) = \frac{3}{{\sqrt {12} }} = \frac{3}{{2\sqrt 3 }} \cr & = \frac{{\sqrt 3 }}{2} = \sin \left( {\theta + {{30}^ \circ }} \right) = \sin {60^ \circ } \cr & \therefore \theta = {30^ \circ } \cr & {\text{co}}{{\text{s}}^2}\theta = {\text{co}}{{\text{s}}^2}{30^ \circ } \cr & = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & = \frac{3}{4} \cr} $$

$$\eqalign{ & 4{\cos ^2}\theta - 4\sqrt 3 \cos \theta + 3 = 0 \cr & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {30^ \circ }{\text{option A}} \cr & 4{\cos ^2}{30^ \circ } - 4\sqrt 3 \cos {30^ \circ } + 3 = 0 \cr & \Rightarrow 4\left( {\frac{3}{4}} \right) - 4\sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right) + 3 = 0 \cr & \Rightarrow 0 = 0 \cr} $$

$$\eqalign{ & {\text{1}}{{\text{1}}^ \circ }{\text{15'}} = {\text{1}}{{\text{1}} + }\frac{{{\text{15}}}}{{60}} = 11 + \frac{1}{4} = \frac{{{{45}^ \circ }}}{4} \cr & {\text{We know }}\pi {\text{ radian}} = {180^ \circ } \cr & {1^ \circ } = \left( {\frac{\pi }{{{{180}^ \circ }}}} \right){\text{radian,}} \cr & \frac{{{{45}^ \circ }}}{4} = \frac{\pi }{{{{180}^ \circ }}} \times \frac{{{{45}^ \circ }}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{{\pi ^c}}}{{16}}{\text{ }} \cr} $$

$$\eqalign{ & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{50}^ \circ } - x} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{cosec}}\left( {{{90}^ \circ } - \left( {{{40}^ \circ } + x} \right)} \right) \cr & {\text{sec}}\left( {4x - {{50}^ \circ }} \right) = {\text{sec}}\left( {{{40}^ \circ } + x} \right) \cr & 4x - {50^ \circ } = {40^ \circ } + x \cr & 3x = {90^ \circ } \cr & x = {30^ \circ } \cr} $$