Question

If $A=B={{60}^{{}^\circ }}$, verify that $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$.

Answer 1

Hint:We know that the given relation is $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$. We will put A and B as ${{60}^{{}^\circ }}$ and then solve LHS and RHS individually and if we get $LHS=RHS$, then the above relation is verified. We will also use the values $\tan {{0}^{{}^\circ }}=0$ , $\tan {{30}^{{}^\circ }}=\dfrac{1}{\sqrt{3}}$, $\tan {{45}^{{}^\circ }}=1$, $\tan {{60}^{{}^\circ }}=\sqrt{3}$ and $\tan {{90}^{{}^\circ }}=\infty $.

Complete step-by-step answer:
It is given in the question that , then we have to verify that $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$. We will solve LHS and RHS individually by putting the value of $A=B={{60}^{{}^\circ }}$ both sides and if both LHS and RHS are equal then the given relation will be verified.
To proceed with the question, let us first consider the below table. It shows the values of the trigonometric ratios at standard angles.

In LHS we have $\tan \left( A-B \right)$, putting the value $A=B={{60}^{{}^\circ }}$, we get
$=\tan \left( {{60}^{{}^\circ }}-{{60}^{{}^\circ }} \right)$ or
$=\tan \left( {{0}^{{}^\circ }} \right)$
We know that that the value of $\tan {{0}^{{}^\circ }}=0$. therefore,
$=0$.
In RHS we have $\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ , we put the given values $A=B={{60}^{{}^\circ }}$, we get
$=\dfrac{\tan {{60}^{{}^\circ }}-\tan {{60}^{{}^\circ }}}{1+\tan {{60}^{{}^\circ }}\tan {{60}^{{}^\circ }}}$, we know that the value of $\tan {{60}^{{}^\circ }}=\sqrt{3}$, therefore putting this value in the RHS expression, we get
$=\dfrac{\sqrt{3}-\sqrt{3}}{1+\sqrt{3}\sqrt{3}}$ , therefore, on solving, we get
$=0$.
Thus, RHS evaluates to zero.
Now comparing LHS and RHS, we find that \[LHS\text{ }=\text{ }RHS=0\] or simply we can say that
$LHS=RHS$.
Hence Proved.

Note: Student may take the value of $\tan {{60}^{{}^\circ }}=\dfrac{1}{\sqrt{3}}$ and value of $\tan {{30}^{{}^\circ }}=\sqrt{3}$ as both these values are confusing and if someone takes the wrong values, solution may be wrong.So, students should remember trigonometric standard angles and formulas to solve these types of questions.