Question

cot(x-y)=cotxcoty+1/cotx-cotx ? please answer

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cot(x-y)}$

$\tt{=\dfrac{cos(x-y)}{sin(x-y)}}$

$\tt{=\dfrac{cos(x)\,cos(y)+sin(x)\,sin(y)}{sin(x)\,cos(y)-cos(x)\,sin(y)}}$

$\tt{=\dfrac{\dfrac{cos(x)\,cos(y)+sin(x)\,sin(y)}{sin(x)\,sin(y)}}{\dfrac{sin(x)\,cos(y)-cos(x)\,sin(y)}{sin(x)\,sin(y)}}}$

$\tt{=\dfrac{\dfrac{cos(x)\,cos(y)}{sin(x)\,sin(y)}+\dfrac{sin(x)\,sin(y)}{sin(x)\,sin(y)}}{\dfrac{sin(x)\,cos(y)}{sin(x)\,sin(y)}-\dfrac{cos(x)\,sin(y)}{sin(x)\,sin(y)}}}$

$\tt{=\dfrac{\dfrac{cos(x)}{sin(x)}\cdot\dfrac{cos(y)}{sin(y)}+1}{\dfrac{cos(y)}{sin(y)}-\dfrac{cos(x)}{sin(x)}}}$

$\tt{=\dfrac{\dfrac{cos(y)}{sin(y)}\cdot\dfrac{cos(x)}{sin(x)}+1}{\dfrac{cos(y)}{sin(y)}-\dfrac{cos(x)}{sin(x)}}}$

$\tt{=\dfrac{cot(y)\cdot\,cot(x)+1}{cot(y)-cot(x)}}$