Question

Prove that $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$?

Answer 1

Hint: We first use the submultiple formulas $1+\cos A=2{{\cos }^{2}}\dfrac{A}{2}$, $1-\cos A=2{{\sin }^{2}}\dfrac{A}{2}$, $\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$ to break the expression. Then we take $2\cos \dfrac{A}{2}$ and $2\sin \dfrac{A}{2}$ common. We omit the $\left( \cos \dfrac{A}{2}-\sin \dfrac{A}{2} \right)$ part and multiply $\cos \dfrac{A}{2}$. The final multiplication gives the right hand part of $\dfrac{1+\cos A}{\sin A}$ on simplification. .

Complete step by step solution:
We first simplify the left-hand side expression of $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}$.
We use the formulas of submultiple $1+\cos A=2{{\cos }^{2}}\dfrac{A}{2}$, $1-\cos A=2{{\sin }^{2}}\dfrac{A}{2}$, $\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$.
We get $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 1+\cos A \right)-\sin A}{\sin A-\left( 1-\cos A \right)}$.
So, $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2{{\cos }^{2}}\dfrac{A}{2}-2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}-2{{\sin }^{2}}\dfrac{A}{2}}$.
Now we take $2\cos \dfrac{A}{2}$ and $2\sin \dfrac{A}{2}$ common from the numerator and the denominator respectively.
$\dfrac{2{{\cos }^{2}}\dfrac{A}{2}-2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}-2{{\sin }^{2}}\dfrac{A}{2}}=\dfrac{2\cos \dfrac{A}{2}\left( \cos \dfrac{A}{2}-\sin \dfrac{A}{2} \right)}{2\sin \dfrac{A}{2}\left( \cos \dfrac{A}{2}-\sin \dfrac{A}{2} \right)}$.
We omit the $\left( \cos \dfrac{A}{2}-\sin \dfrac{A}{2} \right)$ part from both the numerator and the denominator.
We get $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2\cos \dfrac{A}{2}}{2\sin \dfrac{A}{2}}$.
We now multiply $\cos \dfrac{A}{2}$ part to both the numerator and the denominator.
We get $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2\cos \dfrac{A}{2}\cos \dfrac{A}{2}}{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}=\dfrac{2{{\cos }^{2}}\dfrac{A}{2}}{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}$.
We use the submultiple formulas again to get
$\begin{align}
  & \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1} \\
 & =\dfrac{2{{\cos }^{2}}\dfrac{A}{2}}{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}} \\
 & =\dfrac{1+\cos A}{\sin A} \\
\end{align}$
We now break the summation and get $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}=\csc A+\cot A$.
Thus proved $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$.

Note:
We can also simplify the right-hand part through back process solving to reach the common part of $\dfrac{2\cos \dfrac{A}{2}}{2\sin \dfrac{A}{2}}$. As the two parts give the same solution, we can say that both of the expressions are equal.