Question

Determine the value of x, such that $2cose{c}^2{30}^0+xsi{n}^2{60}^0-\frac{3}{4}ta{n}^2{30}^0=10$.

Answer 1

$2cose{c}^2{30}^{\circ }+xsi{n}^2{60}^{\circ }-\frac{3}{4}ta{n}^2{30}^{\circ }=10$

$L.H.S:2(2{)}^2+x(\frac{\sqrt{3}}{2}{)}^2-\frac{3}{4}(\frac{1}{\sqrt{3}}{)}^2=10$ $[\because cosec{30}^{\circ }=2,sin{60}^{\circ }=\frac{\sqrt{3}}{2},tan{30}^{\circ }=\frac{1}{\sqrt{3}}]$

$\Rightarrow 8+\frac{3x}{4}-\frac{3}{4\times 3}=10$

$\Rightarrow 8+\frac{3x}{4}-\frac{1}{4}=10$

$\Rightarrow \frac{32+3x-1}{4}=10$

$\Rightarrow 3x+31=40$

$\Rightarrow 3x=40-31$

$\Rightarrow 3x=9$

$\Rightarrow x=\frac{9}{3}$

$\Rightarrow x=3$