$$\eqalign{ & {\text{According to the question ,}} \cr & {\text{ se}}{{\text{c}}^4}{\text{A}}\left( {1 - {{\sin }^4}{\text{A}}} \right) - 2{\text{ta}}{{\text{n}}^2}{\text{A}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow {\text{ se}}{{\text{c}}^4}{45^ \circ }\left( {1 - {{\sin }^4}{{45}^ \circ }} \right) - 2{\text{ta}}{{\text{n}}^2}{45^ \circ } \cr & \Rightarrow 4\left( {1 - \frac{1}{4}} \right) - 2 \cr & \Rightarrow 4 \times \frac{3}{4} - {\text{2}} \cr & \Rightarrow 3 - 2 \cr & \Rightarrow 1 \cr} $$

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

$$\eqalign{ & \pi \sin \theta = 1\,.....(i) \cr & \pi \cos \theta = 1\,.....(ii) \cr & {\text{Divide eq}}{\text{. (i) from (ii)}} \cr & \frac{{\pi \sin \theta }}{{\pi \cos \theta }} = \frac{1}{1} \cr & \tan \theta = 1 \cr & \tan \theta = \tan {45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore \sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1 \cr & = \sqrt 3 \tan \left( {\frac{2}{3} \times {{45}^ \circ }} \right) + 1 \cr & = \sqrt 3 {\text{ tan}}{30^ \circ } + 1 \cr & = \sqrt 3 \times \frac{1}{{\sqrt 3 }} + 1 \cr & = 2 \cr} $$

$$\eqalign{ & {\text{In }}\vartriangle {\text{, }}\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C}} = {\text{18}}{0^ \circ }\,....{\text{(i)}} \cr & {\text{sin A}} = {\text{cos B}} \cr & {\text{sin A}} = {\text{sin}}\left( {{{90}^ \circ } - {\text{B}}} \right){\text{ }} \cr & {\text{A}} = {90^ \circ } - {\text{B}} \cr & {\text{A}} + {\text{B}} = {90^ \circ }\,........(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \angle {\text{C}} = {90^ \circ } \cr & {\text{So, cos C }} = \cos {90^ \circ } = 0 \cr} $$

$$\eqalign{ & {\bf{Shortcut\,\, method:}} \cr & {\text{Put }}\theta = {30^ \circ } \cr & \Rightarrow \sec \theta - \tan \theta = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \sec {30^ \circ } - \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{{\sqrt 3 }} - \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}{\text{ }}\left( {{\text{Satisfied}}} \right) \cr & \sec \theta = {30^ \circ } \cr & \Rightarrow \sec \theta .\tan \theta \cr & \Rightarrow \sec {30^ \circ }.\tan {30^ \circ } \cr & \Rightarrow \frac{2}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{2}{3} \cr} $$

$$\eqalign{ & \Rightarrow \frac{{\cos \theta }}{{1 - \sin \theta }} + \frac{{\cos \theta }}{{1 + \sin \theta }} = 4 \cr & \Rightarrow \cos \theta \left( {\frac{{1 + \sin \theta + 1 - \sin \theta }}{{1 - {{\sin }^2}\theta }}} \right) = 4 \cr & \Rightarrow \cos \theta \left( {\frac{2}{{{\text{co}}{{\text{s}}^2}\theta }}} \right) = 4 \cr & \Rightarrow \cos \theta = \frac{1}{2} \cr & \Rightarrow \theta = {60^ \circ } \cr} $$

$$\eqalign{ & \sec 15\theta = \operatorname{cosec} 15\theta \cr & \Rightarrow \frac{1}{{\cos {{15} }\theta }} = \frac{1}{{\sin 15\theta }} \cr & \Rightarrow \frac{{\sin 15\theta }}{{{\text{cos15}}\theta }} = 1 \cr & \Rightarrow {\text{tan15}}\theta = 1 \cr & \Rightarrow {\text{tan15}}\theta = {\text{tan}}{45^ \circ } \cr & \Rightarrow 15\theta = {45^ \circ } \cr & \Rightarrow \theta = \frac{{45}}{{15}} \cr & \Rightarrow \theta = {3^ \circ } \cr} $$

$$\eqalign{ & {\text{7co}}{{\text{s}}^2}\theta + 3{\sin ^2}\theta = 4 \cr & 7{\text{co}}{{\text{s}}^2}\theta + 3\left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) - 4 = 0 \cr & 7{\text{co}}{{\text{s}}^2}\theta - 3{\text{co}}{{\text{s}}^2}\theta + 3 - 4 = 0 \cr & 4{\text{co}}{{\text{s}}^2}\theta = 1 \cr & {\text{co}}{{\text{s}}^2}\theta = \frac{1}{4} \cr & {\text{cos}}\theta = \frac{1}{2} \cr & \cos \theta = {\text{cos}}{60^ \circ } \cr & \theta = {60^ \circ } \cr} $$

$${\text{ta}}{{\text{n}}^2}x + {\text{co}}{{\text{t}}^2}x$$
We know that the value of tan²x and cot²x is minimum at 45°
$$\eqalign{ & \therefore {\text{ta}}{{\text{n}}^2}{45^ \circ } + {\text{co}}{{\text{t}}^2}{45^ \circ } \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & \sin \theta - \cos \theta = \frac{7}{{13}} = \alpha \cr & {\text{When, }} \cr & ax + by = m\,......(i) \cr & bx - ay = n\,......(ii) \cr} $$
By adding these two equations after making square on both sides we get,
$$\eqalign{ & \left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) = {m^2} + {n^2} \cr & {\text{In the same process}} \cr & \sin \theta \pm \cos \theta = a \cr & {\text{Then,}}\sin \theta \pm \cos \theta = \sqrt {2 - {a^2}} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {2 - {{\left( {\frac{7}{{13}}} \right)}^2}} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {2 - \left( {\frac{{49}}{{169}}} \right)} \cr & \Rightarrow \sin \theta + \cos \theta = \sqrt {\frac{{289}}{{169}}} \cr & \Rightarrow \sin \theta + \cos \theta = \frac{{17}}{{13}} \cr} $$

$$\eqalign{ & \sin \theta \times \cos \theta = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & 2\sin \theta \times \cos \theta = 1 \cr & \sin 2\theta = 1 \cr & \sin 2\theta = \sin {90^ \circ } \cr & 2\theta = {90^ \circ } \cr & \theta = {45^ \circ } \cr & {\text{So,}}\sin \theta - {\text{cos}}\theta \cr & = \sin {45^ \circ } - \cos {45^ \circ } \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0{\text{ }} \cr} $$

$$\eqalign{ & = {\cos ^2}{20^ \circ } + {\cos ^2}{70^ \circ } \cr & = {\cos ^2}\left( {{{90}^ \circ } - {{70}^ \circ }} \right) + {\cos ^2}{70^ \circ } \cr & = 1 \cr} $$

(cosec a - sin a) (sec a - cos a) (tan a + cot a)
Put a = 45°
⇒ (cosec45° - sin45°) (sec45° - cos45°) (tan45° + cot45°)
$$\eqalign{ & \Rightarrow \left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)\left( {1 + 1} \right) \cr & \Rightarrow \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \times 2 \cr & \Rightarrow \frac{1}{2} \times 2 \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & {\text{Hit and Trial method}} \cr & {\text{Put }}\theta = {60^ \circ }{\text{option A}} \cr & \Rightarrow 2{\sin ^2}{60^ \circ } = 3\cos {60^ \circ } \cr & \Rightarrow 2{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = 3\left( {\frac{1}{2}} \right) \cr & \Rightarrow \frac{3}{2} = \frac{3}{2}{\text{ }}\left( {{\text{LHS}} = {\text{RHS}}} \right) \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0 \cr & \Rightarrow {a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 \cr & \therefore a = b = c \cr & \vartriangle {\text{ABC}} = {\text{equilateral }}\vartriangle \cr & \therefore \angle {\text{A}} = \angle {\text{B}} = \angle {\text{C}} = {60^ \circ } \cr & {\text{So, }}{\sin ^2}{\text{A}} + {\sin ^2}{\text{B}} + {\sin ^2}{\text{C}} \cr & \Rightarrow {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & \Rightarrow 3{\sin ^2}{60^ \circ } \cr & \Rightarrow 3 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4} \cr} $$

$$\eqalign{ & {\text{tan}}\theta = {\text{tan}}{30^ \circ }.{\text{tan}}{60^ \circ } \cr & {\text{tan}}\theta = \frac{1}{{\sqrt 3 }}.\sqrt 3 \cr & {\text{tan}}\theta = 1 \cr & {\text{tan}}\theta = {\text{tan}}{45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore 2\theta = {90^ \circ } \cr} $$

$$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & 4{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{4} \cr & {\text{sin }}\theta = \frac{1}{2} \cr & \sin \theta = \sin {30^ \circ } \cr & \theta = {30^ \circ } \cr & \sec \theta + \operatorname{cosec} \theta \cr & = \sec {30^ \circ } + \operatorname{cosec} {30^ \circ } \cr & = \frac{2}{{\sqrt 3 }} + 2 \cr} $$

$$\eqalign{ & \left( {\frac{{\sin \theta + \sin \phi }}{{\cos \theta + \cos \phi }} + \frac{{\cos \theta \cos \phi }}{{\sin\theta - \sin\phi }}} \right) \cr & {\text{Put }}\theta = {90^ \circ } \cr & \phi = {0^ \circ } \cr & \therefore \left( {\frac{{\sin90^ \circ + \sin0^ \circ }}{{\cos90^ \circ + \cos0^ \circ }} + \frac{{\cos90^ \circ - \cos0^ \circ }}{{\sin90^ \circ - \sin0^ \circ }}} \right) \cr & \Rightarrow \left( {\frac{{1 + 0}}{{0 + 1}} + \frac{{0 - 1}}{{1 - 0}}} \right) \cr & \Rightarrow \left( {1 - 1} \right) \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {\cos ^2}x + {\cos ^4}x = 1 \cr & \Rightarrow {\cos ^4}x = 1 - {\cos ^2}x \cr & \Rightarrow {\cos ^4}x = {\sin ^2}x \cr & \Rightarrow {\cos ^2}x{\cos ^2}x = {\sin ^2}x \cr & \Rightarrow {\text{co}}{{\text{s}}^2}x = {\text{ta}}{{\text{n}}^2}x\,\,\,....(i) \cr & \Rightarrow {\cos ^4}x = {\text{ta}}{{\text{n}}^4}x\,\,\,....(ii) \cr & ta{n^2}x + {\tan ^4}x \cr & {\text{co}}{{\text{s}}^2}x + {\text{co}}{{\text{s}}^4}x = 1 \cr & {\text{ta}}{{\text{n}}^2}x + {\text{ta}}{{\text{n}}^4}x = 1 \cr & \left( {{\text{From eq}}{\text{. (i) and (ii)}}} \right) \cr} $$

Let x = 8cos10°. cos20°. cos40°
Multiply on both side by sin10° and applying formula (2sinθ. cosθ = sin²θ)
⇒ x sin10° = 4 × 2sin10° cos10°. cos20°. cos40°
⇒ x sin10° = 2 × 2sin20°.cos20°. cos40°
⇒ x sin10° = 2 × sin40°. cos40°
⇒ x sin10° = sin80°
⇒ x sin10° = sin(90° - 10°)
⇒ x sin10° = cos10°
then, x = $$\frac{{\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }}}$$
x = cot10°

$$\eqalign{ & = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr & = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given 4}}{\sin ^2}\theta = 3 \cr & {\sin ^2}\theta = \frac{3}{4} \cr & \sin \theta = \frac{{\sqrt 3 }}{2} \cr & \sin \theta = {\text{sin }}{60^ \circ } \cr & \theta = {60^ \circ } \cr & \because \tan \theta - \cot \frac{\theta }{2} \cr & = \tan {60^ \circ } - \cot \frac{{{{60}^ \circ }}}{2} \cr & = \tan {60^ \circ } - \cot {30^ \circ } \cr & = \sqrt 3 - \sqrt 3 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{tan}}\alpha = 2\left( {{\text{given}}} \right) \cr & \therefore \frac{{{\text{cose}}{{\text{c}}^2}\alpha - {\text{se}}{{\text{c}}^2}\alpha }}{{{\text{cose}}{{\text{c}}^2}\alpha + se{c^2}\alpha }} \cr} $$
(Divide by coses²α both in N and D)
$$\eqalign{ & = \frac{{1 - {\text{ta}}{{\text{n}}^2}\alpha }}{{1 + {\text{ta}}{{\text{n}}^2}\alpha }} \cr & = \frac{{1 - {{\left( 2 \right)}^2}}}{{1 + {{\left( 2 \right)}^2}}} \cr & = - \frac{3}{5} \cr} $$

$${\sin ^2}{25^ \circ } + {\sin ^2}{65^ \circ } + {\text{cose}}{{\text{c}}^2}{57^ \circ } - {\text{ta}}{{\text{n}}^2}{33^ \circ }$$
$$ \Rightarrow \left( {{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }} \right) + $$     $$\left( {{\text{se}}{{\text{c}}^2}{{33}^ \circ } - {\text{ta}}{{\text{n}}^2}{{33}^ \circ }} \right)$$
$$\eqalign{ & \Rightarrow 1 + 1 \cr & \Rightarrow 2 \cr & {\bf{Note:}} \cr & {\sin ^2}{65^ \circ } = {\sin ^2}\left( {{{90}^ \circ } - {{25}^ \circ }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{co}}{{\text{s}}^2}{25^ \circ } \cr & {\text{cose}}{{\text{c}}^2}{57^ \circ } = {\operatorname{cosec} ^2}\left( {{{90}^ \circ } - {{33}^ \circ }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{se}}{{\text{c}}^2}{33^ \circ } \cr} $$

$$\eqalign{ & {\text{ tan}}\theta = \frac{4}{3} \cr & \frac{{\sin \theta }}{{\cos \theta }} = \frac{4}{3} \cr & \Rightarrow \frac{{3 \times 4 + 2 \times 3}}{{3 \times 4 - 2 \times 3}} \cr & \Rightarrow \frac{{18}}{6} \cr & \Rightarrow 3 \cr} $$

$$\eqalign{ & {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\text{se}}{{\text{c}}^2}{60^ \circ } - {\text{co}}{{\text{t}}^2}{60^ \circ } + {\text{ta}}{{\text{n}}^2}{30^ \circ } \cr & = {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( 2 \right)^2} - {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = \frac{4}{3} + 4 - \frac{1}{3} + \frac{1}{3} \cr & = \frac{{16}}{3} \cr & = 5\frac{1}{3} \cr} $$

$$\eqalign{ & {\text{4}}\angle {\text{A}} = {\text{3}}\angle {\text{B}} = {\text{12}}\angle {\text{C}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = \frac{1}{4}:\frac{1}{3}:\frac{1}{{12}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 3:4:1 \cr & {\text{Now,}} \cr & 3x + 4x + x = {180^ \circ } \cr & 8x = {180^ \circ } \cr & x = \frac{{{{180}^ \circ }}}{8} \cr & \angle {\text{A}} = 3x = \frac{{{{180}^ \circ }}}{8} \times 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {67.5^ \circ } \cr} $$

$$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right){\text{ = }}\frac{1}{2} \cr & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = {\text{sin }}{30^ \circ } \cr & \theta + {18^ \circ } = {30^ \circ } \cr & \therefore \theta = {12^ \circ } \cr & {\text{We know that,}} \cr & {180^ \circ } = \pi \cr & {12^ \circ } = \frac{\pi }{{{{180}^ \circ }}} \times 12 = \frac{\pi }{{15}} \cr} $$

$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {\cot {{73}^ \circ }{{\cos }^2}{{22}^ \circ } + \frac{1}{{\cot {{17}^ \circ }se{c^2}{{68}^ \circ }}}} \right)$$
$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left[ {\cot \left( {{{90}^ \circ } - {{17}^ \circ }} \right){{\cos }^2}\left( {{{90}^ \circ } - {{68}^ \circ }} \right) + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right]$$
$$\eqalign{ & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {tan{{17}^ \circ }si{n^2}{{68}^ \circ } + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\tan {17^ \circ }\left( {si{n^2}{{68}^ \circ } + {\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow 1\left( 1 \right) \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \sin \theta - \cos \theta = 0 \cr & \sin \theta = \cos \theta \cr & \theta = {45^ \circ } \cr & {\text{Then,}} \cr & sec\theta + {\text{cosec}}\theta \cr & = \sqrt 2 + \sqrt 2 \cr & = 2\sqrt 2 \cr} $$