Question

If $3x\sin {60}^o.{\cos }^2{30}^o=\frac{{\tan }^2{45}^o.\sec {60}^o}{cosec{60}^o.\sin {45}^o}$, then find the value of $x$.

Answer 1

$3x\sin 60{\cos }^{2}30=\frac{{\tan }^{2}45\cdot \sec 60}{cosec60\sin 45}$

$3x\times \frac{\sqrt{3}}{2}\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{(1{)}^2\times (2)}{\left(\frac{2}{\sqrt{3}}\right)\times 1_{/\sqrt{2}}}$

$3x\times \frac{\sqrt{\text{/}3}}{2}$ $\times \frac{3}{4}$ $=\frac{\text{/}2\sqrt{\text{/}3}\times \sqrt{2}}{\text{/}2}$

$x=\frac{8\sqrt{2}}{9}$.