Question

If $\dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos }^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}={{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{30}^{\circ }}$, then x equals,

a)1
b)-1
c)2
d)0

Answer 1

Hint: Here, we have to substitute the values of all trigonometric ratios where, $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}},$ $\tan {{60}^{\circ }}=\sqrt{3},\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and we can obtain the value of $\csc {{30}^{\circ }}$ and $\sec {{45}^{\circ }}$ by the relation:

Complete step-by-step answer:
$\begin{align}

  & \csc {{30}^{\circ }}=\dfrac{1}{\sin {{30}^{\circ }}} \\

 & \sec {{45}^{\circ }}=\dfrac{1}{\cos {{45}^{\circ }}} \\

\end{align}$

Next by squaring all the above values and substituting in the expression $\dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos }^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}={{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{30}^{\circ }}$ will give the value of x.

Here, we are given that $\dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos }^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}={{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{30}^{\circ }}$.

Now, we have to find the value of x.

First consider the expression,

$\dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos }^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}={{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{30}^{\circ }}$

We know the values for $\tan {{60}^{\circ }}$ and $\tan {{30}^{\circ }}$ where,

$\begin{align}

  & \tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}} \\

 & \tan {{60}^{\circ }}=\sqrt{3} \\

\end{align}$

Next by squaring the above values we obtain,

$\begin{align}

  & \Rightarrow {{\tan }^{2}}{{30}^{\circ }}=\dfrac{1}{3} \\

 & \Rightarrow {{\tan }^{2}}{{60}^{\circ }}=3 \\

\end{align}$

By substituting these values in the above expression we get,

$\dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos }^{2}}{{45}^{\circ }}{{\sin

}^{2}}{{60}^{\circ }}}=3-\dfrac{1}{3}$

Next, by taking LCM we get,

$\Rightarrow \dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos

}^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}=\dfrac{3\times 3-1}{3}$

\[\begin{align}

  & \Rightarrow \dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos

}^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}=\dfrac{9-1}{3} \\

 & \Rightarrow \dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos

}^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}=\dfrac{8}{3} \\

\end{align}\]

We know that

$\begin{align}

  & \cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\

 & \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\

\end{align}$

Now, by taking the squares we obtain,

$\begin{align}

  & \Rightarrow {{\cos }^{2}}{{45}^{\circ }}=\dfrac{1}{2} \\

 & \Rightarrow {{\sin }^{2}}{{60}^{\circ }}=\dfrac{3}{4} \\

\end{align}$

Next, by substituting these values in the above expression, we get,

\[\Rightarrow \dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8\times

\dfrac{1}{2}\times {{\dfrac{3}{4}}^{{}}}}=\dfrac{8}{3}\]

By cancellation, we obtain,

\[\Rightarrow \dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{3}=\dfrac{8}{3}\]

Now, by cross multiplication we get the equation,

\[\Rightarrow x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}=\dfrac{24}{3}\]

\[\Rightarrow x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}=8\] ….. (1)

We also have,

$\begin{align}

  & \csc {{30}^{\circ }}=\dfrac{1}{\sin {{30}^{\circ }}} \\

 & \Rightarrow \csc {{30}^{\circ }}=\dfrac{1}{\dfrac{1}{2}} \\

 & \Rightarrow \csc {{30}^{\circ }}=1\times 2 \\

 & \Rightarrow \csc {{30}^{\circ }}=2 \\

\end{align}$

$\begin{align}

  & \sec {{45}^{\circ }}=\dfrac{1}{\cos {{45}^{\circ }}} \\

 & \Rightarrow \sec {{45}^{\circ }}=\dfrac{1}{\dfrac{1}{\sqrt{2}}} \\

 & \Rightarrow \sec {{45}^{\circ }}=1\times \sqrt{2} \\

 & \Rightarrow \sec {{45}^{\circ }}=\sqrt{2} \\

\end{align}$

Next, by squaring we obtain,

$\begin{align}

  & \Rightarrow {{\csc }^{2}}{{30}^{\circ }}=4 \\

 & \Rightarrow {{\sec }^{2}}{{45}^{\circ }}=2 \\

\end{align}$

In the next step, we have to substitute the above values in equation (1), we get:

\[\begin{align}

  & \Rightarrow x\times 4\times 2=8 \\

 & \Rightarrow 8x=8 \\

\end{align}\]

By cross multiplication,

$\Rightarrow x=1$

Therefore, when $\dfrac{x{{\csc }^{2}}{{30}^{\circ }}{{\sec }^{2}}{{45}^{\circ }}}{8{{\cos }^{2}}{{45}^{\circ }}{{\sin }^{2}}{{60}^{\circ }}}={{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{30}^{\circ }}$ the value of $x=1$.

Hence, the correct answer for this question is option (a).


Note: Here, you must be familiar about values of trigonometric ratios which is very essential to solve this problem especially for the angles ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}$ and ${{90}^{\circ }}$and also should be familiar about the relationships between various trigonometric ratios.