Question

$\frac{\cot A+\text{cosec}A-1}{\cot A-\text{cosec}A+1}$ is equal to

• A. $\text{cosec}A+\cot A$
• B. $\sec A+\cot A$
• C. $\text{cosec}A+\tan A$
• D. $\text{cosec}A-\cot A$

Answer 1

The correct option is A $\text{cosec}A+\cot A$

We know that ${\csc }^{2}A-{\cot }^{2}A=1$

Substituting the above value in the given expression $\frac{\cot A+\csc A-1}{\cot A-\csc A+1}$ as shown below:

$\frac{\cot A+\csc A-1}{\cot A-\csc A+1}=\frac{\cot A+\csc A-({\csc }^{2}A-{\cot }^{2}A)}{\cot A-\csc A+1}=\frac{\cot A+\csc A-\left[\right(\csc A-\cot A\left)\right(\csc A+\cot A\left)\right]}{\cot A-\csc A+1}=\frac{(\cot A+\csc A)[1-(\csc A-\cot A\left)\right]}{\cot A-\csc A+1}$

$=\frac{(\cot A+\csc A)(1-\csc A+\cot A)}{\cot A-\csc A+1}=\frac{(\cot A+\csc A)(\cot A-\csc A+1)}{\cot A-\csc A+1}=\cot A+\csc A$

Hence, $\frac{\cot A+\csc A-1}{\cot A-\csc A+1}=\csc A+\cot A$