Question

How to prove \[\cot (A + B) = (\cot A\cot B - 1) \div (\cot A + \cot B)?\]

Answer 1

Hint:We need to remember here the basic rules of trigonometry first, mainly the formula of \[\cot \theta \]. Then we need to remember the formula of \[\cos (A + B)\] and \[\sin (A + B)\]. By applying these formulas, we can easily solve this question.

Complete step by step answer:
To prove: \[\cot (A + B) = (\cot A\cot B - 1) \div (\cot A + \cot B)\]
Proof:
First, we need to take here the LHS and RHS:
LHS \[ = \cot (A + B)\]
RHS \[ = \cot (A\cot B - 1) \div (\cot A + \cot B)\]
Now, we will see the formula of \[\cot \theta \]. So, we know that:
\[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\]
In LHS:
Here, we will try to put the value of \[\cot \theta \] in LHS, and then we will get:
\[ \Rightarrow \cot (A + B) = \dfrac{{\cos (A + B)}}{{\sin (A + B)}}\]
Where, \[\theta = (A + B)\]

Now, according to this equation, we know that:
\[\cos (A + B) = \cos A\cos B - \sin A\sin B\]
We know this from our trigonometric formulas of \[\cos (A + B)\].Similarly, we will see the trigonometric formulas of \[\sin (A + B)\], and we get:
\[\sin (A + B) = \sin A\cos B + \cos A\sin B\]
By putting the values of \[\cos (A + B)\] and \[\sin (A + B)\], we get:
\[ \Rightarrow \cot (A + B) = \dfrac{{\cos A\cos B - \sin A\sin B}}{{\sin A\cos B + \cos A\sin B}}\]

If we notice the RHS part, the part is in terms of \[\cot \] and there is a term \[1\]. We need to get that term \[1\]. If we look carefully then, according to the LHS, the term \[1\] from the RHS part, resembles the term \[\sin A\sin B\] from the LHS part. If in LHS, we divide \[\sin A\sin B\] to all the terms to get the term \[1\], after dividing, we get:
\[ \Rightarrow \cot (A + B) = \dfrac{{\dfrac{{\cos A\cos B}}{{\sin A\sin B}} - \dfrac{{\sin A\sin B}}{{\sin A\sin B}}}}{{\dfrac{{\sin A\cos B}}{{\sin A\sin B}} + \dfrac{{\cos A\sin B}}{{\sin A\sin B}}}}\]

Here, the term \[\sin A\sin B\] gets cancelled from the numerator and denominator, and we get \[1\].
\[ \Rightarrow \cot (A + B) = \dfrac{{\dfrac{{\cos A\cos B}}{{\sin A\sin B}} - 1}}{{\dfrac{{\sin A\cos B}}{{\sin A\sin B}} + \dfrac{{\cos A\sin B}}{{\sin A\sin B}}}}\]

Now, when we simplify LHS by applying the basic formula of \[\cot \theta \], we get:
\[ \Rightarrow \cot (A + B) = \dfrac{{\cot A \cdot \cot B - 1}}{{\dfrac{{\sin A}}{{\sin A}} \cdot \cot B + \cot A \cdot \dfrac{{\sin B}}{{\sin B}}}}\]

After cancelling all the similar terms, we get:
\[ \Rightarrow \cot (A + B) = \dfrac{{\cot A \cdot \cot B - 1}}{{1 \cdot \cot B + \cot A \cdot 1}}\]
\[ \therefore \cot (A + B) = \dfrac{{\cot A \cdot \cot B - 1}}{{\cot B + \cot A}}\]
Therefore, LHS=RHS (proved).Now, we can say that we have proved the given question.

Note:We need to keep a check on both the LHS and RHS while solving. In this question, when we were solving our LHS, we had to check that in what terms our RHS is present, and then according to that we need to change our LHS to prove the question.