Question

# If $\tan A+\tan B=a\ and\ \cot A+\cot B=b$, Prove that $\cot \left( A+B \right)=\dfrac{1}{a}-\dfrac{1}{b}$.

Hint: We will be using the concept of trigonometric identities to solve the problem. We will be using the data given to us that $\cot A+\cot B=b$ to find the value of $\cot A\cot B$. Also, we will be using the formula that, $\cot \left( A+B \right)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}$ to further simplify the question.

Now, we have been given that,

\begin{align} & \tan A+\tan B=a............\left( 1 \right) \\ & \cot A+\cot B=b............\left( 2 \right) \\ \end{align}

Now, we will inverse the equation (2) as,

$\dfrac{1}{\cot A+\cot B}=\dfrac{1}{b}$

Now, we know the trigonometric identity that,

\begin{align} & \cot A=\dfrac{1}{\tan A} \\ & \cot B=\dfrac{1}{\tan B} \\ \end{align}

So, we have,

$\dfrac{1}{\dfrac{1}{\tan A}+\dfrac{1}{\tan B}}=\dfrac{1}{b}$

Now, on simplify by taking LCM we have,

$\dfrac{\tan A\tan B}{\tan A+\tan B}=\dfrac{1}{b}$

Now, from (2) we substitute the value of $\tan A+\tan B$. So, we have,

\begin{align} & \dfrac{\tan A\tan B}{a}=\dfrac{1}{b} \\ & \tan A\tan B=\dfrac{a}{b} \\ \end{align}

Or we can write it by inverting the terms on both sides as,

\begin{align} & \dfrac{1}{\tan A\tan B}=\dfrac{b}{a} \\ & \cot A\cot B=\dfrac{b}{a}..........\left( 3 \right) \\ \end{align}

Now, we know a trigonometric identity that,

$\cot \left( A+B \right)=\dfrac{\cot A\times \cot B-1}{\cot A+\cot B}$

So, using (2) and (3) we have,

\begin{align} & \cot \left( A+B \right)=\dfrac{\dfrac{b}{a}-1}{b} \\ & =\dfrac{1}{a}-\dfrac{1}{b} \\ \end{align}

Hence, proved that,

\begin{align} & \cot \left( A+B \right)=\dfrac{1}{a}-\dfrac{1}{b} \\ & if\ \tan A+\tan B=a \\ & \cot A+\cot B=b \\ \end{align}

Note: To solve these types of questions it is important to note that we have used the data given to us that $\cot A+\cot B=b$ and then used this data to find the value of $\cot \left( A+B \right)$.