Question

Prove that result: $\frac{1}{\csc A-\cot A}-\frac{1}{\sin A}=\frac{1}{\sin A}-\frac{1}{\csc A+\cot A}$

Answer 1

To prove $\frac{1}{\csc A-\cot A}-\frac{1}{\sin A}=\frac{1}{\sin A}-\frac{1}{\csc A+\cot A}$ is equivalent to prove $\frac{1}{\csc A-\cot A}+\frac{1}{\csc A+\cot A}=\frac{1}{\sin A}+\frac{1}{\sin A}$.

Now, $\frac{1}{\csc A-\cot A}+\frac{1}{\csc A+\cot A}=\frac{2\csc A}{{\csc }^{2}A-{\cot }^{2}A}=$ $2\csc A=\frac{2}{\sin A}$. [Since, ${\csc }^{2}A-{\cot }^{2}A=1$]

So, we have proved, the required equality.