Show that: $\dfrac{1-\sin 60{}^\circ }{\cos 60{}^\circ }=\dfrac{\tan 60{}^\circ -1}{\tan 60{}^\circ +1}$ .

Hint: To solve the above question, you need to individually solve the right-hand side and left-hand side of the equation individually by putting the required values and show that both sides of the equation are equal.

Complete step-by-step answer:


Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.


First, let us start with the graph of sinx.




Next, let us see the graph of cosx.



Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $

Let us start the solution to the above question by simplifying the left-hand side of the equation $\dfrac{1-\sin 60{}^\circ }{\cos 60{}^\circ }=\dfrac{\tan 60{}^\circ -1}{\tan 60{}^\circ +1}$ by putting the values $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\cos 60{}^\circ =\dfrac{1}{2}$ .

$\dfrac{1-\sin 60{}^\circ }{\cos 60{}^\circ }$

$=\dfrac{1-\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}$

$=\dfrac{\dfrac{2-\sqrt{3}}{2}}{\dfrac{1}{2}}$

$=2-\sqrt{3}$

So, the left-hand side of the equation $\dfrac{1-\sin 60{}^\circ }{\cos 60{}^\circ }=\dfrac{\tan 60{}^\circ -1}{\tan 60{}^\circ +1}$ is equal to $2-\sqrt{3}$ .

Now let us simplify the right-hand side of the equation by putting the value $\tan 60{}^\circ =\sqrt{3}$ .

$\dfrac{\tan 60{}^\circ -1}{\tan 60{}^\circ +1}$

$=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

Now we will multiply and divide the expression by $\sqrt{3}-1$ . On doing so, we get

$=\dfrac{\left( \sqrt{3}-1 \right)\left( \sqrt{3}-1 \right)}{\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}$

We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ . Therefore, our
expression becomes:

$\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}-1}$

$=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{2}$

We also know that $\left( a-b \right)={{a}^{2}}+{{b}^{2}}-2ab$ . Therefore, our expression becomes:

$\dfrac{{{\left( \sqrt{3} \right)}^{2}}+1-2\sqrt{3}}{2}$

$=\dfrac{3+1-2\sqrt{3}}{2}$

$=\dfrac{4-2\sqrt{3}}{2}$

$=2-\sqrt{3}$

Therefore, we can say that left-hand side of the equation $\dfrac{1-\sin 60{}^\circ }{\cos 60{}^\circ }=\dfrac{\tan 60{}^\circ -1}{\tan 60{}^\circ +1}$ is equal to the right-hand side. So, we have proved that $\dfrac{1-\sin 60{}^\circ }{\cos 60{}^\circ }=\dfrac{\tan 60{}^\circ -1}{\tan 60{}^\circ +1}$ .


Note: Be careful about the calculation and the signs of the formulas you use as the signs in the formulas are very confusing and are very important for solving the problems. Also, it would help if you remember the properties related to complementary angles and trigonometric ratios.