Question

Prove the following:
$\dfrac{1}{\tan 3A-\tan A}-\dfrac{1}{\cot 3A-\cot A}=\cot 2A$.

Answer 1

To solve this problem first convert all the trigonometric functions into tan A. After this conversion, try to make the denominator of two terms same so that operation could be performed easily. Finally apply some suitable identity of tan to obtain the desired result.

Complete step-by-step answer:
According to the problem, we are provided the expression: $\dfrac{1}{\tan 3A-\tan A}-\dfrac{1}{\cot 3A-\cot A}$.
We have to prove the equivalence of the above expression with $\cot 2A$.
To prove this equivalence between expressions, first take the second term of the expression provided. Now, taking -1 common from the denominator and converting the subtraction into addition, we get
$\begin{aligned} & \Rightarrow \dfrac{1}{\tan 3A-\tan A}-\dfrac{1}{-\left( \cot A-\cot 3A \right)} \\\ & \Rightarrow \dfrac{1}{\tan 3A-\tan A}+\dfrac{1}{\cot A-\cot 3A} \\\ \end{aligned}$
Now, converting the cot function into tan function by using the identity: $\cot \theta =\dfrac{1}{\tan \theta }$
$\Rightarrow \dfrac{1}{\tan 3A-\tan A}+\dfrac{1}{\dfrac{1}{\tan A}-\dfrac{1}{\tan 3A}}$
Now, we simplify the denominator of the second term to obtain the denominator which is same as the first term
$\begin{aligned} & \Rightarrow \dfrac{1}{\tan 3A-\tan A}+\dfrac{1}{\dfrac{\tan 3A-\tan A}{\tan A\cdot \tan 3A}} \\\ & \Rightarrow \dfrac{1}{\tan 3A-\tan A}+\dfrac{\tan A\cdot \tan 3A}{\tan 3A-\tan A} \\\ \end{aligned}$
Finally, we get a simplified equation in tan function which is,
$\Rightarrow \dfrac{1+\tan A\cdot \tan 3A}{\tan 3A-\tan A}\ldots (1)$
Now, one useful formula of tan(A-B) can be expressed as:
$\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A\cdot \tan B}$
Now, to apply the above formula taking the reciprocal of equation (1), we get
$\begin{aligned} & \Rightarrow \dfrac{1}{\dfrac{\tan 3A-\tan A}{1+\tan A\cdot \tan 3A}} \\\ & \Rightarrow \dfrac{1}{\tan (3A-A)} \\\ & \Rightarrow \dfrac{1}{\tan 2A}=\cot 2A \\\ \end{aligned}$
So, we proved the equivalence of the expression as the right-hand side is equal to the left-hand side. Hence, we obtain the desired expression as given in the problem statement.

Note: Another way of solving this problem is by expanding the identity of tan 3A. But this method will introduce a large amount of terms in the simplification part and hence it will make our problem complex. Therefore, by using the above methodology students can easily obtain the desired result without errors.