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Question: \(\frac{{\sin h}x - {\sin h}y}{{\cos h}x - {\cos h}y}\) is equal to...

sin⁡hx−sin⁡hycos⁡hx−cos⁡hy\frac{{\sin h}x - {\sin h}y}{{\cos h}x - {\cos h}y} is equal to

A

2coth⁡(x+y)2\coth(x + y)

B

tan⁡h(x+y2){\tan h}\left( \frac{x + y}{2} \right)

C

coth⁡(x+y2)\coth\left( \frac{x + y}{2} \right)

D

cot⁡h(x−y2)\cot h\left( \frac{x - y}{2} \right)

Answer

coth⁡(x+y2)\coth\left( \frac{x + y}{2} \right)

Explanation

Solution

sin⁡hx−sin⁡hycos⁡hx−cos⁡hy=2cos⁡hx+y2sin⁡hx−y22sin⁡hx+y2sin⁡hx−y2=cot⁡h(x+y2)\frac{{\sin h}x - {\sin h}y}{{\cos h}x - {\cos h}y} = \frac{2{\cos h}\frac{x + y}{2}{\sin h}\frac{x - y}{2}}{2{\sin h}\frac{x + y}{2}{\sin h}\frac{x - y}{2}} = {\cot h}\left( \frac{x + y}{2} \right)