Question

Prove the following expression, $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Answer 1

Hint: We will first use the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and then substitute $\left( A+B \right)$ in the place of $\theta $.Then we will use the identities of $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. And finally, we will divide both the numerator and denominator by $\cos A\cos B$ to get the answer.

Complete step by step answer:
In the question, we are asked to prove the given expression, $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. So, to prove this we will consider the left hand side or the LHS of the equation, which is given as $\tan \left( A+B \right)$. Now, we know that there is a trigonometric identity, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ for any value of $\theta $ So, we will take the value of $\theta $ as $\left( A+B \right)$ here. So, by substituting these values, we will get the LHS as,
$\tan \left( A+B \right)=\dfrac{\sin \left( A+B \right)}{\cos \left( A+B \right)}$
Now, we will apply the identities of $\sin \left( A+B \right)$ and $\cos \left( A+B \right)$ which is given by, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ respectively. So, applying these in the above equation, we get,
$\tan \left( A+B \right)=\dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$
Now, we will divide both the numerator and denominator by $\cos A\cos B$. So, we get the above equation as follows,
$\tan \left( A+B \right)=\dfrac{\dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}}{\dfrac{\cos A\cos B-\sin A\sin B}{\cos A\cos B}}$
We know that this can also be written as follows,
$\tan \left( A+B \right)=\dfrac{\dfrac{\sin A\cos B}{\cos A\cos B}+\dfrac{\cos A\sin B}{\cos A\cos B}}{\dfrac{\cos A\cos B}{\cos A\cos B}-\dfrac{\sin A\sin B}{\cos A\cos B}}$
Which on simplification, can be written as follows,
$\tan \left( A+B \right)=\dfrac{\dfrac{\sin A}{\cos A}+\dfrac{\sin B}{\cos B}}{1-\dfrac{\sin A\sin B}{\cos A\cos B}}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, so we will apply that here in the above equation. So, we get,
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
We can see that this is what was given in the right hand side or the RHS of the given expression.
Therefore, we can say that since LHS = RHS, we have proved the expression.

Note: The students should be careful while applying the identities of $\sin \left( A+B \right)$ and $\cos \left( A+B \right)$. They should not write any wrong term or sign while writing that, like, $\sin \left( A+B \right)=\sin A\cos A+\cos B\sin B$ as this could lead to an incorrect answer.