$$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$$

$$ \Rightarrow $$ $$(\cos x + \cos 3x)$$ + $$(\cos 2x + \cos 4x)$$ = 0

$$ \Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0$$

$$ \Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0$$

$$\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0$$

$$x = \pi ,{\pi \over 2},{{3\pi } \over 2},{\pi \over 5},{{3\pi } \over 5},{{7\pi } \over 5},{{9\pi } \over 5}$$

Let $${f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)$$

Consider

$${f_4}\left( x \right) - {f_6}\left( x \right) $$

$$=$$ $${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$$

$$ = {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]$$

$$ = {1 \over 4} - {1 \over 6} = {1 \over {12}}$$

From Sine Rule

$${{AB} \over {\sin \theta }} = {{\sqrt {{p^2} + {q^2}} } \over {\sin \left( {\pi - \left( {\theta + \alpha } \right)} \right)}}$$

$$AB = {{\sqrt {{p^2} + {q^2}} \sin \theta } \over {\sin \theta \cos \alpha + \cos \theta \sin \alpha }}$$

$$ = {{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {q\sin \theta + p\cos \theta }}$$

(As $$\cos \alpha = {q \over {\sqrt {{p^2} + {q^2}} }}$$ and $$\sin \alpha = {p \over {\sqrt {{p^2} + {q^2}} }}$$ )

Given expression can be written as

$${{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}$$

(As $$\tan A = {{\sin A} \over {\cos A}}$$ and $$\cot A = {{\cos A} \over {\sin A}}$$ )

$$ = {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}$$

$$ = {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}$$

$$ = 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$$