Question

$\frac{\sin hx-\sin hy}{\cos hx-\cos hy}=$

• A. $2\cot \left\{h(x+y)\right\}$
• B. $\tan h\left\{\frac{\left[x+y\right]}{2}\right\}$
• C. $\cot h\left\{\frac{\left[x+y\right]}{2}\right\}$
• D. $\cot h\left\{\frac{\left[x-y\right]}{2}\right\}$

Answer 1

The correct option is C

$\cot h\left\{\frac{\left[x+y\right]}{2}\right\}$

Explanation for the correct answer:

Simplifying the given expression using trigonometric identities:

$$\begin{gathered} \frac{\sin hx-\sin hy}{\cos hx-\cos hy} \\ =\frac{2\cos h\left(\frac{x+y}{2}\right)\sin h\left(\frac{x-y}{2}\right)}{2\sin h\left(\frac{x+y}{2}\right)\sin h\left(\frac{x-y}{2}\right)}\left[\because \sin x-\sin y=2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\text{and}\cos x-\cos y=2\sin h\left(\frac{x+y}{2}\right)\sin h\left(\frac{x-y}{2}\right)\right] \\ =\cot h\left(\frac{x+y}{2}\right) \end{gathered}$$

Thus, $\frac{\sin hx-\sin hy}{\cos hx-\cos hy}=$$\cot h\left[\frac{x+y}{2}\right]$

Therefore, the correct answer is option (C).