Respuesta :
The sum and difference identity cos(x + y) / sin(x - y) = 1 - cotxcoty / cotx - coty is verified
Solution:
Given expression is:
[tex]\frac{\cos (x+y)}{\sin (x-y)}=\frac{1-\cot x \cot y}{\cot x-\cot y}[/tex]
Let us first solve L.H.S
[tex]\frac{\cos (x+y)}{\sin (x-y)}[/tex] ------ 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,
[tex]=\frac{\cos x \cos y-\sin x \sin y}{\sin x \cos y-\sin y \cos x}[/tex]
[tex]\text { Taking sinx } \times \text { siny as common }[/tex]
[tex]=\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)}[/tex]
[tex]\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}[/tex]
Taking -1 as common from numerator and denominator we get,
[tex]\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}[/tex]
= R.H.S
Thus L.H.S = R.H.S
Thus the given expression has been verified using sum and difference identity
