 QUESTION

# Prove the following trigonometric identity:$\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\sin A\tan A-\cot A\cos A$

Hint: Take the left hand side or the given expression. Put $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A}$.
Then do the multiplication and arrange the terms to get the right hand side.

We have to prove the following identity:
$\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\sin A\tan A-\cot A\cos A$
Let us first take the left hand side of the above expression.
$\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)$
We know that $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A}$. Put these values in the above expression.
$=\left( 1+\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A} \right)\left( \sin A-\cos A \right)$
By multiplying both the terms we will get,
$=\sin A\left( 1+\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A} \right)-\cos A\left( 1+\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A} \right)$
$=\sin A+\cos A+\dfrac{{{\sin }^{2}}A}{\cos A}-\left( \cos A+\dfrac{{{\cos }^{2}}A}{\sin A}+\sin A \right)$
Now we will adjust terms in such a way so that we can get the right hand side.
$=\sin A+\cos A+\dfrac{\sin A}{\cos A}\times \sin A-\cos A-\sin A-\dfrac{\cos A}{\sin A}\times \cos A$
Now we can cancel out the opposite terms from the above expression and we will put $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A}$
Therefore,
$=\tan A\sin A-\cot A\cos A$, this is our right hand side expression
Hence,
$\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\sin A\tan A-\cot A\cos A$

Note: Alternatively we can start the proof with the right hand side. That is:
$\sin A\tan A-\cot A\cos A$
Put $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A}$.
$=\sin A\dfrac{\sin A}{\cos A}-\cos A\dfrac{\cos A}{\sin A}$
$=\dfrac{{{\sin }^{2}}A}{\cos A}-\dfrac{{{\cos }^{2}}A}{\sin A}$
Now take the left hand side. By putting $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A}$ and multiplying both the terms we will get:
$=\dfrac{{{\sin }^{2}}A}{\cos A}-\dfrac{{{\cos }^{2}}A}{\sin A}$
Therefore,
Left hand side = right hand side.