Question

Prove the following statement.
$\dfrac{\cot A\cos A}{\cot A+\cos A}=\dfrac{\cot A-\cos A}{\cot A\cos A}$

Answer 1

Hint: In order to prove the given relation, we should have some knowledge regarding the concepts of trigonometry, like the trigonometric formulas, $\cot A=\dfrac{\cos A}{\sin A}$ and ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. Also, we should know the algebraic property that, $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. By using these properties, we will get the desired result.
Complete step-by-step answer:
In this question, we have been asked to prove $\dfrac{\cot A\cos A}{\cot A+\cos A}=\dfrac{\cot A-\cos A}{\cot A\cos A}$. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
$LHS=\dfrac{\cot A\cos A}{\cot A+\cos A}$
Now, we know that, $\cot A=\dfrac{\cos A}{\sin A}$. So, we will get the LHS as,
$LHS=\dfrac{\dfrac{\cos A}{\sin A}\cos A}{\dfrac{\cos A}{\sin A}+\cos A}$
We can further write it as,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{\sin A}}{\dfrac{\cos A}{\sin A}+\cos A}$
Now, we will take the LCM of the terms in the denominator. So, we will get,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{\sin A}}{\dfrac{\cos A+\sin A\cos A}{\sin A}}$
We can further write it as,
$LHS=\dfrac{{{\cos }^{2}}A\sin A}{\sin A\left( \cos A+\sin A\cos A \right)}$
We know that common terms of the numerator and the denominator can be cancelled out, so we get,
$LHS=\dfrac{{{\cos }^{2}}A}{\cos A+\sin A\cos A}$
We can see that cos A is common in both the terms of the denominator, so we get,
$LHS=\dfrac{{{\cos }^{2}}A}{\cos A\left( 1+\sin A \right)}$
After cancelling the common term cos A from the numerator and denominator, we will get,
$LHS=\dfrac{\cos A}{\left( 1+\sin A \right)}$
Now, we will multiply the numerator and the denominator by (1 – sin A) to rationalize (1 + sin A). So, we get,
$LHS=\dfrac{\cos A\left( 1-\sin A \right)}{\left( 1+\sin A \right)\left( 1-\sin A \right)}$
Now, we know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. So, we can write $\left( 1+\sin A \right)\left( 1-\sin A \right)=1-{{\sin }^{2}}A$. So, we will get the LHS as,
$LHS=\dfrac{\cos A-\cos A\sin A}{1-{{\sin }^{2}}A}$
And we know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. So, we can write $1-{{\sin }^{2}}A={{\cos }^{2}}A$. Therefore, we get,
$LHS=\dfrac{\cos A-\cos A\sin A}{co{{\operatorname{s}}^{2}}A}$
Now, we will divide the numerator and denominator by sin A. So, we will get,$LHS=\dfrac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{{{\cos }^{2}}A}{\sin A}}$
It can be further written as,
\[\begin{align}
  & LHS=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\cos A\sin A}{\sin A}}{\dfrac{\cos A}{\sin A}\cos A} \\
 & LHS=\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}\cos A} \\
\end{align}\]
We know that, $\cot A=\dfrac{\cos A}{\sin A}$. So, we can write $\dfrac{\cos A}{\sin A}=\cot A$. Therefore, we get,
$\begin{align}
  & LHS=\dfrac{\cot A-\cos A}{\cot A\cos A} \\
 & LHS=RHS \\
\end{align}$
Hence proved.

Note: In this question, the possible mistake one can make is by directly rationalizing the LHS denominator without converting them into cos and sin. This is also correct, but this method will be lengthier and complicated.