Converting Angles To Something Relatable

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

This can be written as

$\dfrac{1*\cos(10^{\circ}) - \sqrt{3}*\sin(10^{\circ})}{\sin(10^{\circ})\cos(10^{\circ})} = \dfrac{4*\left(\dfrac{1}{2}\cos(10^{\circ}) - \dfrac{\sqrt{3}}{2}\sin(10^{\circ})\right)}{2\sin(10^{\circ})\cos(10^{\circ})} =$

$\dfrac{4*(\sin(150^{\circ})\cos(10^{\circ}) + \cos(150^{\circ})\sin(10^{\circ}))}{\sin(20^{\circ})} = \dfrac{4\sin(160^{\circ})}{\sin(20^{\circ})} = \boxed{4}$ ,

since $\sin(x) = \sin(180^{\circ} - x)$ .

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

$\begin{aligned} \frac 1{\sin 10^\circ} - \frac {\sqrt{3}}{\cos 10^\circ} & = \frac {2\times \frac 12}{\sin 10^\circ} - \frac {2 \times \frac {\sqrt{3}}{2}}{\cos 10^\circ} \\ & = 2 \left(\frac {\sin 30^\circ}{\sin 10^\circ} - \frac {\cos 30^\circ}{\cos 10^\circ} \right) \\ & = 2 \left(\frac {3\sin 10^\circ - 4 \sin^3 10^\circ}{\sin 10^\circ} - \frac {4\cos^3 10^\circ - 3 \cos 10^\circ}{\cos 10^\circ} \right) \\ & = 2 \left(3 - 4 \sin^2 10^\circ - 4\cos^2 10^\circ + 3 \right) \\ & = 2 \left(6 - 4 \right) \\ & = \boxed{4} \end{aligned}$

sin10=cos80=cos2(40) =2(cos40)^2-1. =2×(2cos^2(20)-1)^2-1. =2×(4cos^4(20)-4cos^2(20)+1)-1. =8cos^4(20)-8cos^2(20)+2-1 =8cos^2(20){cos^2(20)-1}+1. =8(1-2sin^2(10))^2{(1-2sin^2(10))^2-1}+1. Put sin10=X.above formula will be like this ,x= 8(1-4x^2+4x^4)(1-4x^2+4x^4-1)+1 so 128x^8-256x^6+160x^4-32x^2-x+1=0. by trying & error X=sin10=nearly .173648 so cos10=(1-.173648^2)^.5=.9848078 so the expression =(1÷(.173648))-(√3÷.9848078)=4.000005