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Question: Prove that \(\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A\)?...

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

Explanation

Solution

We first use the submultiple formulas 1+cos⁡A=2cos⁡2A21+\cos A=2{{\cos }^{2}}\dfrac{A}{2}, 1−cos⁡A=2sin⁡2A21-\cos A=2{{\sin }^{2}}\dfrac{A}{2}, sin⁡A=2sin⁡A2cos⁡A2\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2} to break the expression. Then we take 2cos⁡A22\cos \dfrac{A}{2} and 2sin⁡A22\sin \dfrac{A}{2} common. We omit the (cos⁡A2−sin⁡A2)\left( \cos \dfrac{A}{2}-\sin \dfrac{A}{2} \right) part and multiply cos⁡A2\cos \dfrac{A}{2}. The final multiplication gives the right hand part of 1+cos⁡Asin⁡A\dfrac{1+\cos A}{\sin A} on simplification. .

Complete step by step solution:
We first simplify the left-hand side expression of cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}.
We use the formulas of submultiple 1+cos⁡A=2cos⁡2A21+\cos A=2{{\cos }^{2}}\dfrac{A}{2}, 1−cos⁡A=2sin⁡2A21-\cos A=2{{\sin }^{2}}\dfrac{A}{2}, sin⁡A=2sin⁡A2cos⁡A2\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2}.
We get cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=(1+cos⁡A)−sin⁡Asin⁡A−(1−cos⁡A)\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, cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=2cos⁡2A2−2sin⁡A2cos⁡A22sin⁡A2cos⁡A2−2sin⁡2A2\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 2cos⁡A22\cos \dfrac{A}{2} and 2sin⁡A22\sin \dfrac{A}{2} common from the numerator and the denominator respectively.
2cos⁡2A2−2sin⁡A2cos⁡A22sin⁡A2cos⁡A2−2sin⁡2A2=2cos⁡A2(cos⁡A2−sin⁡A2)2sin⁡A2(cos⁡A2−sin⁡A2)\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 (cos⁡A2−sin⁡A2)\left( \cos \dfrac{A}{2}-\sin \dfrac{A}{2} \right) part from both the numerator and the denominator.
We get cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=2cos⁡A22sin⁡A2\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⁡A2\cos \dfrac{A}{2} part to both the numerator and the denominator.
We get cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=2cos⁡A2cos⁡A22sin⁡A2cos⁡A2=2cos⁡2A22sin⁡A2cos⁡A2\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
cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1 =2cos⁡2A22sin⁡A2cos⁡A2 =1+cos⁡Asin⁡A \begin{aligned} & \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{aligned}
We now break the summation and get cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=1sin⁡A+cos⁡Asin⁡A=csc⁡A+cot⁡A\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 cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=csc⁡A+cot⁡A\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 2cos⁡A22sin⁡A2\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.