Question

Verify The Sum & Difference Identity:

Answer 1

The sum and difference identity cos(x + y) / sin(x - y) = 1 - cotxcoty / cotx - coty is verified

Solution:

Given expression is:

$\frac{\cos (x+y)}{\sin (x-y)}=\frac{1-\cot x \cot y}{\cot x-\cot y}$

Let us first solve L.H.S

$\frac{\cos (x+y)}{\sin (x-y)}$    ------ 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,

$=\frac{\cos x \cos y-\sin x \sin y}{\sin x \cos y-\sin y \cos x}$

$\text { Taking sinx } \times \text { siny as common }$

$=\frac{\sin x \sin y\left(\frac{\cos x \cos y}{\sin x \sin y}-1\right)}{\sin x \sin y\left(\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}\right)}$

$\begin{array}{l}{=\frac{\frac{\cos x}{\sin x} \times \frac{\cos y}{\sin y}-1}{\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}}} \\\\ {=\frac{\cot x \times \cot y-1}{\cot y-\cot x}} \\\\ {=\frac{\cot x \cot y-1}{\cot y-\cot x}}\end{array}$

Taking -1 as common from numerator and denominator we get,

$\begin{array}{l}{=\frac{-(1-\cot x \cot y)}{-(\cot x-\cot y)}} \\\\ {=\frac{(1-\cot x \cot y)}{(\cot x-\cot y)}}\end{array}$

= R.H.S

Thus L.H.S = R.H.S

Thus the given expression has been verified using sum and difference identity