$$\eqalign{ & {\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}; \cr & {\text{Period}} = \frac{{2\pi }}{2} = \pi \cr} $$

$$\sin n\theta = {b_0} + {b_1}\sin \theta + {b_2}{\sin ^2}\theta + .....\,\,\,.$$
This is possible when $$n$$ is an odd integer.
Put $$\theta = 0$$  to get $${b_0}.$$ After differentiating w.r.t. $$\theta ,$$ put $$\theta = 0$$  to get $${b_1}.$$

$$\eqalign{ & \sin \frac{\pi }{{2n}} + \cos \frac{\pi }{{2n}} = \frac{{\sqrt n }}{2} \cr & \Rightarrow \,\,{\sin ^2}\frac{\pi }{{2n}} + {\cos ^2}\frac{\pi }{{2n}} + 2\sin \frac{\pi }{{2n}}\cos \frac{\pi }{{2n}} = \frac{n}{4} \cr & \Rightarrow \,\,1 + \sin \frac{\pi }{4} = \frac{n}{4} \cr & \Rightarrow \,\,\sin \frac{\pi }{4} = \frac{{n - 4}}{4} \cr} $$
For $$n = 2$$  the given equation is not satisfied.
Considering $$n > 1$$  and $$n \ne 2$$
$$\eqalign{ & 0 < \sin \frac{\pi }{4} < 1 \cr & \Rightarrow \,\,0 < \frac{{n - 4}}{4} < 1 \cr & \Rightarrow \,\,4 < n < 8. \cr} $$

Given expression can be written as
$$\eqalign{ & \frac{{\sin A}}{{\cos A}} \times \frac{{\sin A}}{{\sin A - \cos A}} + \frac{{\cos A}}{{\sin A}} \times \frac{{\cos A}}{{\cos A - \sin A}} \cr & \left( {\because \,\,\tan A = \frac{{\sin A}}{{\cos A}}\,{\text{and }}\cot A = \frac{{\cos A}}{{\sin A}}} \right) \cr & = \frac{1}{{\sin A - \cos A}}\left\{ {\frac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A}}} \right\} \cr & = \frac{{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}A}}{{\sin A\cos A}} \cr & = 1 + \sec A\,{\text{cosec}}\,A \cr} $$

Value $$ = \cos {12^ \circ } \cdot \cos {24^ \circ } \cdot \cos {48^ \circ } \cdot \cos \left( {{{180}^ \circ } - {{96}^ \circ }} \right) \cdot \cos {36^ \circ } \cdot \cos {72^ \circ }$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {\cos {{12}^ \circ } \cdot \cos {{24}^ \circ } \cdot \cos {{48}^ \circ } \cdot \cos {{96}^ \circ }} \right)\left( {\cos {{36}^ \circ } \cdot \cos {{72}^ \circ }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - \left( {\frac{{2\sin{{12}^ \circ } \cdot \cos {{12}^ \circ } \cdot \cos {{24}^ \circ } \cdot \cos {{48}^ \circ } \cdot \cos {{96}^ \circ }}}{{2\sin{{12}^ \circ }}}} \right) \times \left( {\frac{{2\sin{{36}^ \circ } \cdot \cos {{36}^ \circ } \cdot \cos {{72}^ \circ }}}{{2\sin{{36}^ \circ }}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = - \frac{{\sin {{192}^ \circ }}}{{{2^4} \cdot \sin {{12}^ \circ }}} \cdot \frac{{\sin {{144}^ \circ }}}{{{2^2}\sin {{36}^ \circ }}} = \frac{1}{{{2^4}}} \cdot \frac{1}{{{2^2}}}. \cr} $$

$$\eqalign{ & p = \sqrt {{a^2} + {b^2}} \left\{ {\frac{a}{{\sqrt {{a^2} + {b^2}} }}\cos \theta - \frac{b}{{\sqrt {{a^2} + {b^2}} }}\sin \theta } \right\} \cr & p = \sqrt {{a^2} + {b^2}} \cos \left( {\theta + \alpha } \right),\,\,{\text{where }}\cos \alpha = \frac{a}{{\sqrt {{a^2} + {b^2}} }}. \cr & {\text{But }} - 1 \leqslant \cos \left( {\theta + \alpha } \right) \leqslant 1. \cr} $$

We have,
$$\eqalign{ & mn\, = \left( {{\text{cosec}}\,\theta - \sin \,\theta } \right)\left( {\sec \,\theta - \cos \,\theta } \right) \cr & = \,\left( {\frac{1}{{\sin \,\theta }} - \sin \,\theta } \right)\left( {\frac{1}{{\cos \,\theta }} - \cos \,\theta } \right) \cr & = \,\frac{{1 - {{\sin }^2}\theta }}{{\sin \,\theta }} \times \frac{{1 - {{\cos }^2}\theta }}{{\cos \,\theta }} \cr & = \,\frac{{{{\cos }^2}\theta }}{{\sin \,\theta }} \times \frac{{{{\sin }^2}\theta }}{{\cos \,\theta }} = \sin \,\theta \cdot \cos \,\theta \cr & \therefore {m^{\frac{2}{3}}} + {n^{\frac{2}{3}}} = {\left( {\frac{{{{\cos }^2}\theta }}{{\sin \,\theta }}} \right)^{\frac{2}{3}}} + {\left( {\frac{{{{\sin }^2}\theta }}{{\cos \,\theta }}} \right)^{\frac{2}{3}}} \cr & = \frac{{{{\cos }^{\frac{4}{3}}}\theta }}{{{{\sin }^{\frac{2}{3}}}\theta }} + \frac{{{{\sin }^{\frac{4}{3}}}\theta }}{{{{\cos }^{\frac{2}{3}}}\theta }} = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\left( {\sin \,\theta \cdot \cos \,\theta } \right)}^{\frac{2}{3}}}}} \cr & = \,\frac{1}{{{{\left( {mn} \right)}^{\frac{2}{3}}}}} = {\left( {mn} \right)^{ - \frac{2}{3}}} \cr} $$

$$\eqalign{ & {\left( {\cos {{20}^ \circ } + \sin {{20}^ \circ }} \right)^2} + {\left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)^2} = 2 \cr & \therefore \,\,\,\cos {20^ \circ } + \sin {20^ \circ } = \sqrt {2 - {p^2}} > 0 \cr & \therefore \,\,\cos {40^ \circ } = \left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)\left( {\cos {{20}^ \circ } + \sin {{20}^ \circ }} \right) = p\sqrt {2 - {p^2}} . \cr} $$

$$\eqalign{ & {\text{Line}}\,\,y = \sqrt 3 \,\,{\text{and graph}} = y = \tan x \cr & {\text{Now,}}\,{\text{we have }}\sqrt 3 = \tan x \cr & \Rightarrow \tan x = \tan {60^ \circ } \cr & \Rightarrow x = {60^ \circ }\,\,\,\left[ {\because x \in \left( {0,\frac{\pi }{2}} \right)} \right] \cr} $$
Hence, one intersecting point is possible in the given domain i.e., $$k = 1.$$

We have
$$\eqalign{ & 5\,{\tan ^2}x - 5\,{\cos ^2}x = 2\left( {{2\cos^2} 2x - 1} \right) + 9 \cr & \Rightarrow \,\,5\,{\tan ^2}x - 5\,{\cos ^2}x = 4\,{\cos ^2}x - 2 + 9 \cr & \Rightarrow \,\,5\,{\tan ^2}x = 9\,{\cos ^2}x + 7 \cr & \Rightarrow \,\,5\left( {{{\sec }^2}x - 1} \right) = 9\,{\cos ^2}x + 7 \cr & {\text{Let }}{\cos ^2}x = t \cr & \Rightarrow \,\,\frac{5}{t} - 9t - 12 = 0 \cr & \Rightarrow \,\,9{t^2} + 12t - 5 = 0 \cr & \Rightarrow \,\,9{t^2} + 15t - 3t - 5 = 0 \cr & \Rightarrow \,\,\left( {3t - 1} \right)\left( {3t + 5} \right) = 0 \cr & \Rightarrow \,\,t = \frac{1}{3}\,\,{\text{as }}\,t \ne - \frac{5}{3}. \cr & cos2x = 2co{s^2}x - 1 = 2\left( {\frac{1}{3}} \right) - 1 \cr & = - \frac{1}{3} \cr & \cos 4x = 2{\cos ^2}2x - 1 = 2{\left( { - \frac{1}{3}} \right)^2} - 1 \cr & = - \frac{7}{9} \cr} $$