Question

Prove the following statement: $\dfrac{\cot A+\tan B}{\cot B+\tan A}=\cot A\tan B$

Answer 1

Hint: First we are going to write tan in terms of cot and then we are going to prove the above statement by using some trigonometric formulas and then we will start from LHS and then by using the formulas we will show that it is equal to RHS.

Complete step-by-step answer:

Let’s first state some formula that we are going to use to solve this question.

$\cot A=\dfrac{1}{\tan A}$


Now we will solve LHS using this formula,


$\begin{align}


  & \dfrac{\cot A+\tan B}{\cot B+\tan A} \\


 & =\dfrac{\cot A+\dfrac{1}{\cot B}}{\cot B+\dfrac{1}{\cot A}} \\


 & =\dfrac{\dfrac{\cot A\cot B+1}{\cot B}}{\dfrac{\cot A\cot B+1}{\cot A}} \\


 & =\dfrac{\cot A}{\cot B} \\


 & =\cot A\tan B \\


\end{align}$


Hence we have shown using the above formula that LHS = RHS.

Hence Proved.

Note: It’s always better that we check if the answer that we have got by using the above formula is correct or not to avoid some calculation mistake and for that we need to put some values in place of A and B to check whether it satisfies the above expression or not. There are a bunch of trigonometric formulas that should be kept in mind while solving these questions and if we use some different set of formulas then that will be another method to solve this question, but at some point we can see that they are nearly equal. There are many ways to solve this question as one can start from RHS and then using some of the trigonometric formula one can show that RHS = LHS.