Question

If an equation is given by ${{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\sin {{45}^{\circ }}\cos {{45}^{\circ }}$ , then x equals,
(a) 2
(b) -2
(c) $\dfrac{-1}{2}$
(d) $\dfrac{1}{2}$

Answer 1

Hint : Here, first we have to substitute the values of $\sin {{45}^{\circ }}$ and $\cos {{45}^{\circ }}$ where $\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ . Then, we will have the RHS in terms of x. Next, put the values for ${{\tan }^{2}}{{45}^{\circ }}$ and ${{\cos }^{2}}{{30}^{\circ }}$ where $\tan {{45}^{\circ }}=1$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ . Then do the simplification to obtain the value of x.

Complete step-by-step answer :
Here, we are given that ${{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\sin {{45}^{\circ }}\cos {{45}^{\circ }}$ .
Now, we have to find the value of x.
First, let us consider the functions in the RHS of the given expression,
${{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\sin {{45}^{\circ }}\cos {{45}^{\circ }}$
We know the values for $\sin {{45}^{\circ }}$ and $\cos {{45}^{\circ }}$ where,
$\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$
By substituting these values in the RHS of the given expression we get,
$\begin{aligned} & \Rightarrow {{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}} \\\ & \Rightarrow {{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\times \dfrac{1}{2} \\\ \end{aligned}$
In order to keep the variable x on one side, we will perform cross multiplication and then we will obtain the equation as,
$\Rightarrow 2({{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }})=x.....\left( i \right)$
We know the values for $\tan {{45}^{\circ }}$ and $\cos {{30}^{\circ }}$ , so we can compute the values of ${{\tan }^{2}}{{45}^{\circ }}\text{ and }{{\cos }^{2}}{{30}^{\circ }}$ by squaring them. So, we will do it as below,
$\begin{aligned} & \tan {{45}^{\circ }}=1 \\\ & \Rightarrow {{\tan }^{2}}{{45}^{\circ }}={{\left( 1 \right)}^{2}} \\\ & \Rightarrow {{\tan }^{2}}{{45}^{\circ }}=1 \\\ \end{aligned}$
$\begin{aligned} & \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow {{\cos }^{2}}{{30}^{\circ }}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\\ & \Rightarrow {{\cos }^{2}}{{30}^{\circ }}=\dfrac{3}{4} \\\ \end{aligned}$
Now, by substituting the values of ${{\tan }^{2}}{{45}^{\circ }}$ and ${{\cos }^{2}}{{30}^{\circ }}$ in equation (i) we get:
$\Rightarrow 2\left( 1-\dfrac{3}{4} \right)=x$
Now, by taking LCM we obtain the equation,
$\begin{aligned} & \Rightarrow 2\left( \dfrac{4-3}{4} \right)=x \\\ & \Rightarrow 2\left( \dfrac{1}{4} \right)=x \\\ & \Rightarrow 2\times \dfrac{1}{4}=x \\\ \end{aligned}$
Next, by cancellation, we get the value of x,
$\Rightarrow x=\dfrac{1}{2}$
Therefore, we can say that for ${{\tan }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}=x\sin {{45}^{\circ }}\cos {{45}^{\circ }}$ , the value of $x=\dfrac{1}{2}$ .
Hence, the correct answer to this question is option (d).

Note : Here, you must be familiar with values of trigonometric ratios which is very essential to solve this problem especially for all the standard angles ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}$ and ${{90}^{\circ }}$ . Some students try to substitute the given options in the given equation to check which one satisfies the equation. But, this approach is going to be a waste of time. In any case, we have to find the values of all the trigonometric functions first, then simplify to get the RHS and LHS in order to check if it is satisfied.