Question

Show that $\cot \left( {a + b} \right) = \dfrac{{\cot a\cot b - 1}}{{\cot a + \cot b}}$.

Answer 1

To prove that $\cot \left( {a + b} \right) = \dfrac{{\cot a\cot b - 1}}{{\cot a + \cot b}}$, we will be taking LHS. Now, we can write $\cot \left( {a + b} \right)$ as $\dfrac{{\cos \left( {a + b} \right)}}{{\sin \left( {a + b} \right)}}$. Now, the formulas for $\cos \left( {a + b} \right)$ and $\sin \left( {a + b} \right)$ are
$\Rightarrow \cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
$\Rightarrow \sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$
Using these formulas and some calculations, we will get LHS=RHS.

Complete step-by-step answer:
In this question, we are given a trigonometric identity and we need to prove it.
Given identity: $\cot \left( {a + b} \right) = \dfrac{{\cot a\cot b - 1}}{{\cot a + \cot b}}$ - - - - - - - - - - (1)
This is a general formula we will be using to solve many trigonometric questions. But in this question, we are going to see how this formula was derived.
Let us take the LHS of the equation (1) and try to bring it as the RHS.
So the LHS of the equation (1) is $\cot \left( {a + b} \right)$.
$\Rightarrow LHS = \cot \left( {a + b} \right)$- - - - - - - - - - - - (2)
Now, we know that cot is cos divided by sin. Hence, we can write $\cot \left( {a + b} \right)$ as $\cos \left( {a + b} \right)$ divided by $\sin \left( {a + b} \right)$. Therefore, equation (2) becomes
$\Rightarrow LHS = \cot \left( {a + b} \right) = \dfrac{{\cos \left( {a + b} \right)}}{{\sin \left( {a + b} \right)}}$- - - - - - - (3)
Now, we know that the formula for $\cos \left( {a + b} \right)$ is
$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
and the formula for $\sin \left( {a + b} \right)$ is
$\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$
Hence, equation (3) becomes
$\Rightarrow LHS = \dfrac{{\cos \left( {a + b} \right)}}{{\sin \left( {a + b} \right)}} = \dfrac{{\cos a\cos b - \sin a\sin b}}{{\sin a\cos b + \cos a\sin b}}$- - - - - - (4)
Now, dividing equation (4) with $\sin a\sin b$, we get
$\Rightarrow LHS = \dfrac{{\dfrac{{\cos a\cos b}}{{\sin a\sin b}} - \dfrac{{\sin a\sin b}}{{\sin a\sin b}}}}{{\dfrac{{\sin a\cos b}}{{\sin a\sin b}} + \dfrac{{\cos a\sin b}}{{\sin a\sin b}}}}$
Now, $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$ and $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$
$\Rightarrow LHS = \dfrac{{\cot a\cot b - 1}}{{\cot b + \cot a}} \\\ \Rightarrow LHS = \dfrac{{\cot a\cot b - 1}}{{\cot a + \cot b}} \\\$
Which is the RHS of the equation (1).
$\Rightarrow LHS = RHS$
Hence, we have proved that $\cot \left( {a + b} \right) = \dfrac{{\cot a\cot b - 1}}{{\cot a + \cot b}}$.

Note: We can also prove $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$ using the same method.
$\Rightarrow LHS = \tan \left( {a + b} \right) = \dfrac{{\sin \left( {a + b} \right)}}{{\cos \left( {a + b} \right)}}$
$\Rightarrow LHS = \dfrac{{\sin a\cos b + \cos a\sin b}}{{\cos a\cos b - \sin a\sin b}}$
Divide the above equation with $\cos a\cos b$, we get
$\Rightarrow LHS = \dfrac{{\dfrac{{\sin a\cos b}}{{\cos a\cos b}} + \dfrac{{\cos a\sin b}}{{\cos a\cos b}}}}{{\dfrac{{\cos a\cos b}}{{\cos a\cos b}} - \dfrac{{\sin a\sin b}}{{\cos a\cos b}}}} \\\ \Rightarrow LHS = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}} \\\ \Rightarrow LHS = RHS \\\$