Question

If the value of $A=B={{60}^{\circ }}$, verify that
(i) $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
(ii) $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
(iii) $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

Answer 1

Hint: We solve this question by substitution method, place the values of $A=B={{60}^{\circ }}$ in the equations and by using standard trigonometric angles simplify and verify whether L.H.S=R.H.S.

Complete step-by-step answer:
We will consider the part (i) given by $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B...(i)$
Now, we will do the verification of $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ by substitution. We will first consider the left hand side of the expression which is $\cos \left( A-B \right)$. Here, we will do the substitution by placing the degree = ${{60}^{\circ }}$ to the angles A and B.
$\begin{align}
  & \Rightarrow \cos \left( A-B \right)=\cos \left( {{60}^{\circ }}-{{60}^{\circ }} \right) \\
 & \Rightarrow \cos \left( {{60}^{\circ }}-{{60}^{\circ }} \right)=\cos \left( 0 \right) \\
\end{align}$
As the value of $\cos \left( 0 \right)=1$.
$\Rightarrow \cos \left( A-B \right)=1$
Now, we will consider the right hand side of the expression (i) which is given by $\cos A\cos B+\sin A\sin B$ substituting the degree = ${{60}^{\circ }}$ to the angles A and B.
$\Rightarrow \cos A\cos B+\sin A\sin B=\cos \left( {{60}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)+\sin \left( {{60}^{\circ }} \right)\sin \left( {{60}^{\circ }} \right)$
As the value of $\cos \left( {{60}^{\circ }} \right)=\dfrac{1}{2}$ and $\sin \left( {{60}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$.
$\begin{align}
  & \Rightarrow \cos A\cos B+\sin A\sin B=\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \cos A\cos B+\sin A\sin B=\dfrac{1}{4}+ \dfrac{3}{4} \\
 & \Rightarrow \cos A\cos B+\sin A\sin B=\dfrac{4}{4} \\
 & \Rightarrow \cos A\cos B+\sin A\sin B=1 \\
\end{align}$
Since, clearly, the expression has its left hand side equal to the right hand side which is equal to 1. So, expression (i) is verified.
Now, we will consider the (ii) expression given by $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B...(ii)$.
We will first consider the left hand side of the expression $\sin \left( A-B \right)$ and substitute the values of degree = ${{60}^{\circ }}$ in the angles A and B.
$\begin{align}
  & \Rightarrow \sin \left( A-B \right)=\sin \left( {{60}^{\circ }}-{{60}^{\circ }} \right) \\
 & \Rightarrow \sin \left( A-B \right)=\sin \left( {{60}^{\circ }} \right) \\
\end{align}$
As,
 $\begin{align}
  & \sin \left( 0 \right)=0 \\
 & \Rightarrow \sin \left( A-B \right)=0 \\
\end{align}$
Now, we will consider the right hand side of the expression (ii) and substitute the value of degree = ${{60}^{\circ }}$in the angles A and B.
$\Rightarrow \sin A\cos B-\cos A\sin B=\sin \left( {{60}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)-\cos \left( {{60}^{\circ }} \right)\sin \left( {{60}^{\circ }} \right)$
As the value of $\sin \left( {{60}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ and $\cos \left( {{60}^{\circ }} \right)=\dfrac{1}{2}$.
$\begin{align}
  & \Rightarrow \sin A\cos B-\cos A\sin B=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \sin A\cos B-\cos A\sin B=0 \\
\end{align}$
So, we have clearly that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$. Now, we will consider the third expression which is given by $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}...(iii)$.
We will first consider the left hand side of the expression which is $\tan \left( A-B \right)$. Now, we will substitute $A={{60}^{\circ }}$ and $B={{60}^{\circ }}$. Thus, we have $\tan \left( A-B \right)=\tan \left( {{60}^{\circ }}-{{60}^{\circ }} \right)$ which is $\tan \left( A-B \right)=\tan \left( 0 \right)$. Since, $\tan \left( 0 \right)$ has the value 0.
$\Rightarrow \tan \left( A-B \right)=0$.
Now, we will consider the right hand side of the expression which is given by the following and substitute $A={{60}^{\circ }}$ and $B={{60}^{\circ }}$.
$\Rightarrow \dfrac{\tan A-\tan B}{1+\tan A\tan B}=\dfrac{\tan \left( {{60}^{\circ }} \right)-\tan \left( {{60}^{\circ }} \right)}{1+\tan \left( {{60}^{\circ }} \right)\tan \left( {{60}^{\circ }} \right)}$
As the value of $\tan \left( {{60}^{\circ }} \right)=\sqrt{3}$.
$\begin{align}
  & \Rightarrow \dfrac{\tan A-\tan B}{1+\tan A\tan B}=\dfrac{\sqrt{3}-\sqrt{3}}{1+\left( \sqrt{3}\sqrt{3} \right)} \\
 & \Rightarrow \dfrac{\tan A-\tan B}{1+\tan A\tan B}=0 \\
\end{align}$
The left and right hand side of the third expression are equal. So, we have that the equation is satisfied.
Hence, all the equations are satisfied.

Note: The verification should be done after substituting the values of trigonometric terms. It will be better to consider the left hand side of the expression first and then consider the right hand side of the expression separately instead of verifying them with including the equal sign. For example, the following way of verifying will not be considered,
$\begin{align}
  & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \\
 & \Rightarrow \tan \left( {{60}^{\circ }}-{{60}^{\circ }} \right)=\dfrac{\tan \left( {{60}^{\circ }} \right)-\tan \left( {{60}^{\circ }} \right)}{1+\tan \left( {{60}^{\circ }} \right)\tan \left( {{60}^{\circ }} \right)} \\
 & \Rightarrow \tan \left( 0 \right)=\dfrac{\sqrt{3}-\sqrt{3}}{1+\left( \sqrt{3}\sqrt{3} \right)} \\
 & \Rightarrow 0=0 \\
\end{align}$