Question

# Prove that $\cot {\text{A cot3A + cot2A cot3A - cot2A cotA = - 1 }}$

Hint: Here in this question we will use the property of cot (A+B) to solve the problem.

Now, we know the property of tan (A+B) but we can’t remember the property of cot (A+B). So, to overcome this problem we will derive the property of cot from the property of tan.
Now, tan (A+B) = $\dfrac{{\tan {\text{A + tanB}}}}{{1 - \tan {\text{A tanB}}}}$. Also, we know that $\tan {\text{ x = }}\dfrac{1}{{\cot {\text{ x}}}}$. So, putting value of tan in the property of tan (A+B), we get
$\dfrac{1}{{\cot ({\text{A + B)}}}} = \dfrac{{\dfrac{1}{{\cot {\text{A}}}} + \dfrac{1}{{\cot {\text{B}}}}}}{{1 - \dfrac{1}{{\cot {\text{A cotB}}}}}}$
Simplifying the above term, we get
$\cot ({\text{A + B) = }}\dfrac{{{\text{cotA cotB - 1}}}}{{\cot {\text{A + cot B}}}}$ ……….. (1)
Now, to solve the given question, put A = 2A and B = A in the equation (1)
$\cot (2{\text{A + A) = }}\dfrac{{\cot 2{\text{A cotA - 1}}}}{{{\text{cot2A + }}{\text{ cotA}}}}$
By cross – multiplying both sides, we get
$\cot 3{\text{A ( cot2A + cotA) = cot2A cotA - 1}}$
$\cot 3{\text{A cot2A + cot3A cotA = cot2A cotA - 1}}$
Rearranging the terms in the above equation,
$\cot {\text{A cot3A + cot2A cot3A - cot2A cotA = - 1 }}$
Hence proved.

Note: In the questions which include the property of cot there are many mistakes done by students in solving the problems. Most of the students don't know the property of cot correctly and apply the incorrect formula resulting in the wrong answer. Easiest way to remove such error is that you should derive the property of cot from the property of tan which is as comparison to cot is easy to remember.