Question

please solve its as you can

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{x {cosec}^{2}30 \degree \: {sec}^{2} 45\degree }{8 {cos}^{2} 45\degree {sin}^{2} 60\degree } = {tan}^{2}60\degree - {tan}^{2}30\degree$

We know,

$\rm :\longmapsto\:cosec30\degree = 2$

$\rm :\longmapsto\:sec45\degree = \sqrt{2}$

$\rm :\longmapsto\:cos45\degree = \dfrac{1}{ \sqrt{2} }$

$\rm :\longmapsto\:sin60\degree = \dfrac{ \sqrt{3} }{2}$

$\rm :\longmapsto\:tan60\degree = \sqrt{3}$

$\rm :\longmapsto\:tan30\degree = \dfrac{1}{ \sqrt{3} }$

So, on substituting these values in above expression, we get

$\rm :\longmapsto\:\dfrac{x {(2)}^{2} {( \sqrt{2}) }^{2} }{8 \times {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2} \times {\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }^{2} } = {( \sqrt{3} )}^{2} - {\bigg(\dfrac{1}{ \sqrt{3} } \bigg) }^{2}$

$\rm :\longmapsto\:\dfrac{x \times 4 \times 2}{8 \times \dfrac{1}{2} \times \dfrac{3}{4} } = 3 - \dfrac{1}{3}$

$\rm :\longmapsto\:\dfrac{x }{ \dfrac{3}{8}} = \dfrac{9 - 1}{3}$

$\rm :\longmapsto\:\dfrac{8x }{3} = \dfrac{8}{3}$

$\bf\implies \:x = 1$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$