Answer:
what do I do again
Step-by-step explanation:
To find the width you would divide 203.2 by 80. You answer is 2.54 inches.
Hope this helps!!!
It is 5 hun and 74 your welcome
The sum and difference identity cos(x + y) / sin(x - y) = 1 - cotxcoty / cotx - coty is verified
<u>Solution:</u>
Given expression is:
![\frac{\cos (x+y)}{\sin (x-y)}=\frac{1-\cot x \cot y}{\cot x-\cot y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ccos%20%28x%2By%29%7D%7B%5Csin%20%28x-y%29%7D%3D%5Cfrac%7B1-%5Ccot%20x%20%5Ccot%20y%7D%7B%5Ccot%20x-%5Ccot%20y%7D)
Let us first solve L.H.S
------ EQN 1
We have to use the sum and difference formulas
cos(A + B) = cosAcosB – sinAsinB
sin(A - B) = sinAcosB – cosAsinB
Applying this in eqn 1 we get,
![=\frac{\cos x \cos y-\sin x \sin y}{\sin x \cos y-\sin y \cos x}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ccos%20x%20%5Ccos%20y-%5Csin%20x%20%5Csin%20y%7D%7B%5Csin%20x%20%5Ccos%20y-%5Csin%20y%20%5Ccos%20x%7D)
![\text { Taking sinx } \times \text { siny as common }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Taking%20sinx%20%7D%20%5Ctimes%20%5Ctext%20%7B%20siny%20as%20common%20%7D)
![=\frac{\sin x \sin y\left(\frac{\cos x \cos y}{\sin x \sin y}-1\right)}{\sin x \sin y\left(\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}\right)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Csin%20x%20%5Csin%20y%5Cleft%28%5Cfrac%7B%5Ccos%20x%20%5Ccos%20y%7D%7B%5Csin%20x%20%5Csin%20y%7D-1%5Cright%29%7D%7B%5Csin%20x%20%5Csin%20y%5Cleft%28%5Cfrac%7B%5Ccos%20y%7D%7B%5Csin%20y%7D-%5Cfrac%7B%5Ccos%20x%7D%7B%5Csin%20x%7D%5Cright%29%7D)
![\begin{array}{l}{=\frac{\frac{\cos x}{\sin x} \times \frac{\cos y}{\sin y}-1}{\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}}} \\\\ {=\frac{\cot x \times \cot y-1}{\cot y-\cot x}} \\\\ {=\frac{\cot x \cot y-1}{\cot y-\cot x}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%3D%5Cfrac%7B%5Cfrac%7B%5Ccos%20x%7D%7B%5Csin%20x%7D%20%5Ctimes%20%5Cfrac%7B%5Ccos%20y%7D%7B%5Csin%20y%7D-1%7D%7B%5Cfrac%7B%5Ccos%20y%7D%7B%5Csin%20y%7D-%5Cfrac%7B%5Ccos%20x%7D%7B%5Csin%20x%7D%7D%7D%20%5C%5C%5C%5C%20%7B%3D%5Cfrac%7B%5Ccot%20x%20%5Ctimes%20%5Ccot%20y-1%7D%7B%5Ccot%20y-%5Ccot%20x%7D%7D%20%5C%5C%5C%5C%20%7B%3D%5Cfrac%7B%5Ccot%20x%20%5Ccot%20y-1%7D%7B%5Ccot%20y-%5Ccot%20x%7D%7D%5Cend%7Barray%7D)
Taking -1 as common from numerator and denominator we get,
![\begin{array}{l}{=\frac{-(1-\cot x \cot y)}{-(\cot x-\cot y)}} \\\\ {=\frac{(1-\cot x \cot y)}{(\cot x-\cot y)}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%3D%5Cfrac%7B-%281-%5Ccot%20x%20%5Ccot%20y%29%7D%7B-%28%5Ccot%20x-%5Ccot%20y%29%7D%7D%20%5C%5C%5C%5C%20%7B%3D%5Cfrac%7B%281-%5Ccot%20x%20%5Ccot%20y%29%7D%7B%28%5Ccot%20x-%5Ccot%20y%29%7D%7D%5Cend%7Barray%7D)
= R.H.S
Thus L.H.S = R.H.S
Thus the given expression has been verified using sum and difference identity