Question

If $x\tan {{45}^{\circ }}\cos {{60}^{\circ }}=\sin {{60}^{\circ }}\cot {{60}^{\circ }}$ then x is equal to:
(a) 1
(b) $\sqrt3$
(c) $\dfrac{1}{2}$
(d) $\dfrac{1}{\sqrt{2}}$

Answer 1

Hint:In the given equation, substitute the different angles of tan, cot, sin and cosine values and then using simple algebra find the value of x. From trigonometry, the values of these trigonometric functions are $\tan {{45}^{\circ }}=1,\cos {{60}^{\circ }}=\dfrac{1}{2},\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2},\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}}$ .

Complete step-by-step answer:
The equation in which we have to find the value of x is:
$x\tan {{45}^{\circ }}\cos {{60}^{\circ }}=\sin {{60}^{\circ }}\cot {{60}^{\circ }}$
To solve the above equation, we need the values of trigonometric functions which we are going to show below:
$\begin{aligned} & \tan {{45}^{\circ }}=1,\cos {{60}^{\circ }}=\dfrac{1}{2} \\\ & \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2},\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$
Now, substituting the above trigonometric function values in the given equation we get,
$x\left( 1 \right)\left( \dfrac{1}{2} \right)=\dfrac{\sqrt{3}}{2}\left( \dfrac{1}{\sqrt{3}} \right)$
In the above equation $\sqrt{3}$ will be cancelled out from the numerator and denominator of the expression written in the L.H.S of the above equation.
$x\left( \dfrac{1}{2} \right)=\dfrac{1}{2}$
In the above equation $\dfrac{1}{2}$ will be cancelled out on both the sides of the equation and we get,
$x=1$
From the above calculations, we are getting the value of x as 1.
Hence, the correct option is (a).

Note: The alternate method of solving the above problem is as follows:
The equation given in the question which contains x:
$x\tan {{45}^{\circ }}\cos {{60}^{\circ }}=\sin {{60}^{\circ }}\cot {{60}^{\circ }}$
We know from the trigonometric function values that,
$\cot {{60}^{\circ }}=\dfrac{\cos {{60}^{\circ }}}{\sin {{60}^{\circ }}}$
Substituting the above value of $\cot {{60}^{0}}$ in the given equation we get,
$x\tan {{45}^{\circ }}\cos {{60}^{\circ }}=\sin {{60}^{\circ }}\dfrac{\cos {{60}^{\circ }}}{\sin {{60}^{\circ }}}$
If you carefully look at the above equation then you will find that on the L.H.S of the above equation $\sin {{60}^{0}}$ will be cancelled out from both the numerator and denominator.
$x\tan {{45}^{\circ }}\cos {{60}^{\circ }}=\cos {{60}^{\circ }}$
Now $\cos {{60}^{0}}$ is cancelling out from both the sides then the above equation will look like:
$x\tan {{45}^{\circ }}=1$
We know that $\tan {{45}^{0}}=1$ so plugging this value in the above equation we get,
$x=1$
Hence, we have got the value of $x=1$ from this alternate method.