Question

no one has been able to help with this and get the right answer

Answer 1

The given expression is:

$\frac{1-\cot y}{\sin y-\cos y}$

It is required to simplify to a single trigonometry function using sin y and cos y.

To do this, use trigonometry identity:

$\cot y=\frac{\cos y}{\sin y}$

Hence, the expression can be written as:

$\begin{gathered} \frac{1-\cot y}{\sin y-\cos y}=\frac{1-\frac{\cos y}{\sin y}}{\sin y-\cos y} \\ Rewrite\text{ 1 as }\frac{\sin y}{\sin y}\text{:} \\ =\frac{\frac{\sin y}{\sin y}-\frac{\cos(y)}{\sin(y)}}{\sin(y)-\cos(y)} \end{gathered}$

Simplify the numerator:

$\frac{\frac{\sin y-\cos(y)}{\sin(y)}}{\sin(y)-\cos(y)}$

Simplify the expression further:

$\frac{\sin y-\cos y}{\sin y(\sin y-\cos y)}$

Cancel out common factors in the denominator and numerator:

$\frac{\cancel{\sin(y)-\cos(y)}}{\sin(y)(\cancel{\sin(y)-\cos(y)})}=\frac{1}{\sin(y)}$

Hence, the required answer is 1/sin(y).