Question

Trigonometry - Verify Identity: cos (x - y) / sin (x + y) = 1 + cotxcoty / cotx + coty

Answer 1

Answer:

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

$Left\ side = Right\ side$, hence the identity is verified.

Step-by-step explanation:

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

Working on right hand side.

$\frac{1+cot(x)cot(y)}{cot(x)+cot(y)}$

Substituting $[cot(x)=\frac{cos(x)}{sin(x)}]$ and $[cot(y)=\frac{cos(y)}{sin(y)}]$

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

Taking LCD and adding fractions.

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

Cancelling out the common denominators.

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

Applying sum and difference formulas $[cos(x-y)=cos(x)cos(y)-sin(x)sin(y)][sin(x+y)=sin(x)cos(y)+sin(y)cos(x)]$

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

Left side

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

∵ $Left\ side = Right\ side$, hence the identity is verified.