Question

If $\frac{1}{cos{290}^{\circ }}+\frac{1}{\sqrt{3}sin{250}^{\circ }}=\lambda$, then the value of $3{\lambda }^2-13$ must be ___

Answer 1

$\frac{1}{2}+\frac{1}{\sqrt{3}sin250}=\lambda ,3{\lambda }^2-13=?$

$cos290=cos(270+20)=sin20[coso+vein{3}^{rd}quad]$

$sin250=sin(270-20)=-cos20[sino-vein{3}^{rd}quad]$

$\frac{1}{sin20}+\frac{1}{\sqrt{30}(-cos20)}=\lambda$

$\Rightarrow \frac{1}{sin20}+\frac{1}{\sqrt{3}(-cos20)}=\lambda$

$\Rightarrow 2\frac{(\frac{\sqrt{3}}{2}cos20-\frac{1}{2}sin20)}{\frac{\sqrt{3}}{2}\times \left(2sin20cos20\right)}=\lambda$

$\Rightarrow \frac{2sin60cos20-cos60sin20}{sin60\left(sin40\right)}=\lambda$

$\Rightarrow \frac{2sin(60-20)}{sin60sin40}=\lambda =\frac{2sin40}{sin60sin40}$

$\Rightarrow \lambda =\frac{2}{5_{\frac{3}{2}}}=\frac{4}{\sqrt{3}}$

$3{\lambda }^2-13=3\left(\frac{16}{3}\right)-13$

$=\begin{array}{c}3\end{array}$