Question

Show that $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\cos A+\sin A$.

Answer 1

Hint:We have sum of two terms in the left-hand side of $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\cos A+\sin A$. We multiply $\cos A$ with $\dfrac{\cos A}{1-\tan A}$ and $\sin A$ with $\dfrac{\sin A}{1-\cot A}$ on both numerator and denominator. Then we add them as the denominators are the same. Then we use the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor the numerator. We also need to mention the conditions.

Complete step by step answer:
We multiply $\cos A$ to the numerator and denominator of $\dfrac{\cos A}{1-\tan A}$.
We know $\tan A=\dfrac{\sin A}{\cos A}$. This gives $\tan A.\cos A=\sin A$.So,
$\dfrac{\cos A}{1-\tan A}\times \dfrac{\cos A}{\cos A}=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}$
We multiply $\sin A$ to the numerator and denominator of $\dfrac{\sin A}{1-\cot A}$.
We know $\cot A=\dfrac{\cos A}{\sin A}$. This gives $\cot A.\sin A=\cos A$.
So, $\dfrac{\sin A}{1-\cot A}\times \dfrac{\sin A}{\sin A}=\dfrac{{{\sin }^{2}}A}{\sin A-\cos A}$.

The equation becomes $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}+\dfrac{{{\sin }^{2}}A}{\sin A-\cos A}$.
To make the denominators same we take a negative sign common.
$\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}-\dfrac{{{\sin }^{2}}A}{\cos A-\sin A}$.
The addition gives $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\cos A-\sin A}$
The numerator is in the form of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We now apply the identity theorem for the term ${{\cos }^{2}}A-{{\sin }^{2}}A$. We assume the values $a=\cos A,b=\sin A$.

Applying the theorem, we get ${{\cos }^{2}}A-{{\sin }^{2}}A=\left( \cos A+\sin A \right)\left( \cos A-\sin A \right)$.
The equation becomes $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{\left( \cos A+\sin A \right)\left( \cos A-\sin A \right)}{\left( \cos A-\sin A \right)}$.
We can now eliminate the $\left( \cos A-\sin A \right)$ from both denominator and numerator.
The equation becomes $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\cos A+\sin A$.
Thus verified $\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\left( \cos A+\sin A \right)$.

Note: It is important to remember that the condition to eliminate the $\left( \cos A-\sin A \right)$ from both denominator and numerator is $\left( \cos A-\sin A \right)\ne 0$. No domain is given for the variable $A$. The value of $\tan A\ne 0$ is essential. The simplified condition will be $A\ne n\pi ,n\in \mathbb{Z}$. We also have the multiple angle theorem of ${{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A$.