Trigonometry — Part 1 MCQs with Explanations

Let x = 2sin 2 θ + 3cos 2 θ ⇒ x = 2sin 2 θ + 2cos 2 θ + cos 2 θ ⇒ x = 2(sin 2 θ + cos 2 θ) + cos 2 θ ⇒ x = 2 + cos 2 θ [since sin 2 θ + cos 2 θ = 1] Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2 Alternative Solution: 2sin 2 θ + 3cos 2 θ Minimum value is 2, [If x sin 2 θ + y cos 2 θ, If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]

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

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

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

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

$$\eqalign{ & \cos \pi x = {x^2} - x + \frac{5}{4} \cr & = {x^2} - 2 \times x \times \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + \frac{5}{4} \cr & = {\left( {x - \frac{1}{2}} \right)^2} + 1 > 1 \cr & = - 1 \leqslant \cos x \leqslant 1 \cr} $$ ∴ So, value of x is none of the above

$$\sin \frac{{\pi x}}{2} = {x^2} - 2x + 2$$ 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°

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

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

$$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = 1 - {\cos ^2}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta .{\cos ^2}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^4}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr} $$

$$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$ Divide numerator & denominator by cosθ $$\eqalign{ & = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr & = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr & {\text{Put value of tan}}\theta \cr & = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr & = \frac{6}{2} \cr & = 3 \cr} $$

(secA - cosA) 2 + (cosecA - sinA) 2 - (cotA - tanA) 2 = (sec 2 A + cos 2 A - 2secA.cosA) + (coses 2 A + sin 2 A - 2cosecA.sinA) - (cot 2 A + tan 2 A - 2cotA.tanA) = sec 2 A - tan 2 A + cos 2 A + sin 2 A + coses 2 A - cot 2 A - 2 = 3 - 2 = 1 Alternate shortcut method: (secA - cosA) 2 + (cosecA - sinA) 2 - (cotA - tanA) 2 Put θ = 45° = (sec45° - cos45°) 2 + (cosec45° - sin45°) 2 - (cot45° - tan45°) 2 $$\eqalign{ & = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2} \cr & = \frac{1}{2} + \frac{1}{2} - 0 \cr & = 1 \cr} $$

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

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

$$\eqalign{ & \frac{{\cot {{30}^ \circ } - \cot {{75}^ \circ }}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = \frac{{\tan {{60}^ \circ } - \tan {{15}^ \circ }}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = \frac{{ - \left( {\tan {{15}^ \circ } - \tan {{60}^ \circ }} \right)}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = - 1 \cr} $$

$$\eqalign{ & \left( {{\text{se}}{{\text{c}}^4}\theta - {\text{ta}}{{\text{n}}^4}\theta } \right) \cr & \Rightarrow \left( {{\text{se}}{{\text{c}}^2}\theta - {\text{ta}}{{\text{n}}^2}\theta } \right)\left( {{\text{se}}{{\text{c}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow 1 \times \left( {{\text{se}}{{\text{c}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right)[1 + {\text{ta}}{{\text{n}}^2}\theta = {\text{se}}{{\text{c}}^2}\theta ] \cr & \Rightarrow 1 \times \frac{7}{{12}} \cr & \Rightarrow \frac{7}{{12}} \cr} $$

$$\eqalign{ & \frac{5}{{{\text{se}}{{\text{c}}^2}\theta }}{\text{ + }}\frac{2}{{1 + {\text{co}}{{\text{t}}^2}\theta }}{\text{ + 3}}{\sin ^2}\theta \cr & = 5{\cos ^2}\theta + \frac{2}{{{\text{cose}}{{\text{c}}^2}\theta }} + 3{\sin ^2}\theta \cr & = 5{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta + 3{\sin ^2}\theta \cr & = 5\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) \cr & \left( {\because {{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1} \right) \cr & = 5 \times 1 \cr & = 5 \cr} $$

$$\eqalign{ & \left( {\frac{1}{{\cos \theta }} + \frac{1}{{\cot \theta }}} \right){\text{ }}\left( {\frac{1}{{\cos \theta }} - \frac{1}{{\cot \theta }}} \right) \cr & = \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) \cr & = {\sec ^2}\theta - {\tan ^2}\theta \left[ {1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right] \cr & = 1 \cr} $$

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

According to the question, $$2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}{67^ \circ } - {\sin ^2}{23^ \circ } - $$ $${\sin ^2}{67^ \circ } - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$ $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$ $${\sin ^2}{23^ \circ } - $$ $${\sin ^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$ $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ ta}}{{\text{n}}^2}{23^ \circ } - $$ $$\left( {{{\sin }^2}{{23}^ \circ } + co{s^2}{{23}^ \circ }} \right) - $$ $${\text{co}}{{\text{t}}^2}{67^ \circ }$$ $$\eqalign{ & \Rightarrow \frac{2}{{{\text{co}}{{\text{s}}^2}{{23}^ \circ }}} - 1 - {\text{co}}{{\text{t}}^2}{67^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{ta}}{{\text{n}}^2}{23^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - \left( {{\text{1}} + {\text{ta}}{{\text{n}}^2}{{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\sec ^2}{23^ \circ } \cr & \Rightarrow se{c^2}{23^ \circ } \cr} $$

$$\eqalign{ & \frac{{{\text{tan}}\theta }}{{1 - {\text{cot}}\theta }} + \frac{{{\text{cot}}\theta }}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{tan}}\theta }}{{1 - \frac{1}{{\tan \theta }}}} + \frac{{\frac{1}{{\tan \theta }}}}{{1 - {\text{tan}}\theta }} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta }}{{{\text{tan}}\theta - 1}} + \frac{1}{{{\text{tan}}\theta \left( {1 - \tan \theta } \right)}} \cr & \Rightarrow \frac{{{\text{tan}}^2\theta }}{{{\text{tan}}\theta - 1}} - \frac{1}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^3}\theta - 1}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{\left( {{\text{tan}}\theta - 1} \right)\left( {{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}} \right)}}{{{\text{tan}}\theta \left( {\tan \theta - 1} \right)}} \cr & \Rightarrow \frac{{{\text{ta}}{{\text{n}}^2}\theta + {\text{tan}}\theta + {\text{1}}}}{{{\text{tan}}\theta }} \cr & \Rightarrow {\text{tan}}\theta + \cot \theta + {\text{1}} \cr} $$

$$\eqalign{ & {\bf{Shortcut \,\, method:}} \cr & = \left( {1 + \cot \theta - \operatorname{cosec} \theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) \cr & \left[ {put,\theta = {{45}^ \circ }} \right] \cr} $$ $$ = \left( {1 + \cot {{45}^ \circ } - \operatorname{cosec} {{45}^ \circ }} \right)$$ $$\left( {1 + \tan {{45}^ \circ } + \sec {{45}^ \circ }} \right)$$ $$\eqalign{ & = \left( {1 + 1 - \sqrt 2 } \right)\left( {1 + 1 + \sqrt 2 } \right) \cr & = \left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right) \cr & = \left[ {{2^2} - {{\left( {\sqrt 2 } \right)}^2}} \right]\left[ {\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \cr & = 4 - 2 \cr & = 2 \cr} $$

$$\eqalign{ & x = a{\text{ }}\sec \theta .\cos \phi \cr & y = b{\text{ }}\sec \theta .sin\phi \cr & z = c{\text{ tan}}\theta \cr & \frac{x}{a} = \sec \theta .\cos \phi \cr & \frac{y}{b} = \sec \theta .sin\phi \cr & \frac{z}{c} = {\text{tan}}\theta \cr & \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr & \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

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

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

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

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

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

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

$$\eqalign{ & {\text{ta}}{{\text{n}}^2}\theta = 1 - {e^2} \cr & \therefore sec\theta + {\text{ta}}{{\text{n}}^3}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .{\text{tan}}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .\frac{{\sin \theta }}{{{\text{cos}}\theta }}.\frac{1}{{\sin \theta }} \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .sec\theta \cr & \Rightarrow sec\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow \sqrt {1 + {\text{ta}}{{\text{n}}^2}\theta } .\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow {\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {1 + 1 - {e^2}} \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {2 - {e^2}} \right)^{\frac{3}{2}}} \cr} $$

$$\eqalign{ & \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a{\text{ }} \cr & \Rightarrow \cos {\text{ }}\alpha = a{\text{ }}\cos {\text{ }}\beta \cr & {\text{On squaring both sides}} \cr & {\cos ^2}\alpha = {a^2}{\cos ^2}\beta \cr & \Rightarrow 1 - {\sin ^2}\alpha = {a^2}\left( {1 - {{\sin }^2}\beta } \right)....(i) \cr & {\text{Again, }}\sin \alpha = {\text{ }}b\sin \beta \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\sin ^2}\alpha = {\text{ }}{b^2}{\sin ^2}\beta \cr & {\text{Put the value of }}{\sin ^2}\alpha {\text{ in equation (i)}} \cr & \Rightarrow {\text{1}} - {b^2}{\sin ^2}\beta = {a^2} - {a^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = {a^2}si{n^2}\beta - {b^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = si{n^2}\beta \left( {{a^2} - {b^2}} \right) \cr & \Rightarrow si{n^2}\beta = \frac{{{a^2} - 1}}{{{a^2} - {b^2}}} \cr} $$

$$\eqalign{ & \sin \theta = 0.7 \cr & \Rightarrow {\sin ^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\left( {0.7} \right)^2} + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow 0.49 + {\text{co}}{{\text{s}}^2}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = 1 - 0.49 \cr & \Rightarrow {\text{cos}}\theta = \sqrt {0.51} \cr} $$

$$\eqalign{ & {\text{2}}\sin \theta + {\text{cos}}\theta = \frac{7}{3} \cr & \Rightarrow \left( {{\text{ta}}{{\text{n}}^2}\theta - {\text{se}}{{\text{c}}^2}\theta } \right) \cr & \Rightarrow \left( {{{\sec }^2}\theta - 1 - {\text{se}}{{\text{c}}^2}\theta } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\left[ {\because 1 + {\text{ta}}{{\text{n}}^2}\theta = {{\sec }^2}\theta } \right] \cr & \Rightarrow - 1 \cr} $$

$$\eqalign{ & {\sec ^2}\theta - \frac{{{{\sin }^2}\theta - 2{{\sin }^4}\theta }}{{{\text{2co}}{{\text{s}}^4}\theta - {\text{co}}{{\text{s}}^2}\theta }} \cr & \Rightarrow {\sec ^2}\theta - \frac{{{{\sin }^2}\theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{{\text{co}}{{\text{s}}^2}\theta \left( {{\text{2co}}{{\text{s}}^2}\theta - 1} \right)}} \cr & \left[ {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta = 2{\text{co}}{{\text{s}}^2}\theta - 1 = 1 - 2{{\sin }^2}\theta } \right] \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \sqrt {\frac{{1 + {\text{sin }}\theta }}{{1 - {\text{sin }}\theta }}} {\text{ + }}\sqrt {\frac{{1 - {\text{sin }}\theta }}{{1 + {\text{sin }}\theta }}} \cr & = \frac{{{{\left( {\sqrt {1 + {\text{sin }}\theta } } \right)}^2} + {{\left( {\sqrt {1 - {\text{sin }}\theta } } \right)}^2}}}{{\sqrt {1 - {{\sin }^2}\theta } }} \cr & = \frac{{1 + {\text{sin }}\theta + {\text{1}} - {\text{sin }}\theta }}{{{\text{cos}}\theta }} \cr & = \frac{2}{{{\text{cos}}\theta }} \cr & = 2{\text{ sec}}\theta \cr} $$ 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° $$\eqalign{ & {\text{ = 2 sec 3}}{0^ \circ } \cr & = \frac{{2 \times 2}}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$

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

∠A + ∠C = ∠B + ∠D = 180° ∴ ∠A = 180° - ∠C cosA = cos(180° - C) ⇒ -cosC Similarly, cosB = -cosD ⇒ cosA + cosB + cosC + cosD ⇒ cosA + cosB - cosA - cosB = 0 Alternate Solution : Put, A = B = C = D = 90° = cosA + cosB + cosC + cosD = cos90° + cos90° + cos90° + cos90° = 0 + 0 + 0 + 0 = 0

$$\eqalign{ & {\text{Given, }}\frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = {\text{2}}\frac{{51}}{{79}} \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta }}{{{\text{sec}}\theta - {\text{tan}}\theta }} = \frac{{209}}{{79}} \cr & {\text{By componendo dividendo}} \cr & \left[ {\frac{a}{b} = \frac{c}{d},{\text{ }}\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}} \right] \cr & \Rightarrow \frac{{{\text{sec}}\theta + {\text{tan}}\theta + {\text{sec}}\theta - {\text{tan}}\theta }}{{{\text{sec}}\theta + {\text{tan}}\theta - {\text{sec}}\theta + {\text{tan}}\theta }} = \frac{{209 + 79}}{{209 - 79}} \cr & \Rightarrow \frac{{2sec\theta }}{{2{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{sec\theta }}{{{\text{tan}}\theta }} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{{\frac{1}{{{\text{cos}}\theta }}}}{{\frac{{\sin \theta }}{{{\text{cos}}\theta }}}} = \frac{{288}}{{130}} \cr & \Rightarrow \frac{1}{{\sin \theta }} = \frac{{288}}{{130}} \cr & \Rightarrow {\text{Therefore, }}\sin \theta = \frac{{130}}{{288}} \cr & \Rightarrow \sin \theta = \frac{{65}}{{144}} \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {\text{1}} + {\text{co}}{{\text{s}}^2}\theta = {\text{3}}\sin \theta .{\text{cos}}\theta {\text{ }}\left( {0 < \theta < \frac{\pi }{2}} \right) \cr & {\text{1}} + {\text{co}}{{\text{s}}^2}\theta = {\text{3}}\sin \theta .{\text{cos}}\theta \cr & {\text{Dividing by }}{\sin ^2}\theta {\text{ in both sides}} \cr & \Rightarrow \frac{{1 + {\text{co}}{{\text{s}}^2}\theta }}{{{{\sin }^2}\theta }} = \frac{{3\sin \theta .{\text{cos}}\theta }}{{{{\sin }^2}\theta }} \cr & \Rightarrow {\text{cose}}{{\text{c}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 3{\text{cot}}\theta \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}\theta + {\text{co}}{{\text{t}}^2}\theta = 3\cot \theta \cr & \left[ {\because 1 + {\text{co}}{{\text{t}}^2}\theta = {\text{cose}}{{\text{c}}^2}\theta } \right] \cr & \Rightarrow 1 + 2{\text{co}}{{\text{t}}^2}\theta = 3\cot \theta \cr & \Rightarrow 2{\text{co}}{{\text{t}}^2}\theta = 3\cot \theta - 1 \cr & {\text{Let }}\theta = {45^ \circ } \cr & \because {\text{cot4}}{{\text{5}}^ \circ } = 1 \cr & \Rightarrow 2{\text{co}}{{\text{t}}^2}{45^ \circ } - 3{\text{cot}}{45^ \circ } + 1 = 0 \cr & \Rightarrow 2 - 3 + 1 = 0 \cr & \Rightarrow 0 = 0 \cr & {\text{Therefore cot}}\theta = {\text{cot}}{45^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \cr} $$