Question

If $\tan (\theta +i\varphi )=\sin (x+iy)$, then the value of $\cot h\left(y\right)\sin h\left(2\varphi \right)$ is

• A. $\tan \left(x\right)\cot \left(2\theta \right)$
• B. $\tan \left(x\right)\sin \left(2\theta \right)$
• C. $\cot \left(x\right)\sin \left(2\theta \right)$
• D. none of these

Answer 1

The correct option is C

$\cot \left(x\right)\sin \left(2\theta \right)$

Explanation for correct option

$\Rightarrow \frac{\sin (\theta +i\varphi )}{\cos (\theta +i\varphi )}=\sin \left(x\right)\cos \left(iy\right)+\sin \left(iy\right)\cos \left(x\right)\because \left[\sin (a+b)=\sin \left(a\right)\cos \left(b\right)+\cos \left(a\right)\sin \left(b\right)\right]$

multiplying $\cos (\theta -i\varphi )$ to both numerator and denominator

$$\begin{gathered} \Rightarrow \frac{\sin (\theta +i\varphi )}{\cos (\theta +i\varphi )}\times \frac{\cos (\theta -i\varphi )}{\cos (\theta -i\varphi )}=\sin \left(x\right)\cos h\left(y\right)+i\sin h\left(y\right)\cos \left(x\right) \\ \Rightarrow \frac{2\sin (\theta +i\varphi )\cos (\theta -i\varphi )}{2\cos (\theta +i\varphi )\cos (\theta -i\varphi )}=\sin \left(x\right)\cos h\left(y\right)+i\sin h\left(y\right)\cos \left(x\right) \\ \Rightarrow \frac{\sin \left(2\theta \right)+\sin \left(2i\varphi \right)}{\cos \left(2\theta \right)+\cos \left(2i\varphi \right)}=\sin \left(x\right)\cos h\left(y\right)+i\sin h\left(y\right)\cos \left(x\right)\because \left[2\sin (a+b)\cos (a-b)=\sin \left(2a\right)+\sin \left(2b\right)\right] \\ \Rightarrow \frac{\sin \left(2\theta \right)}{\cos \left(2\theta \right)+\cos \left(2i\varphi \right)}+\frac{i\sin h\left(2\varphi \right)}{\cos \left(2\theta \right)+\cos \left(2i\varphi \right)}=\sin \left(x\right)\cos h\left(y\right)+i\sin h\left(y\right)\cos \left(x\right) \end{gathered}$$

comparing both sides

$\frac{\sin \left(2\theta \right)}{\cos \left(2\theta \right)+\cos \left(2i\varphi \right)}=\sin \left(x\right)\cos h\left(y\right)$ ---------- (1)

$\frac{\sin h\left(2\varphi \right)}{\cos \left(2\theta \right)+\cos \left(2i\varphi \right)}=\sin h\left(y\right)\cos \left(x\right)$ ----------- (2)

dividing both equations

$$\begin{gathered} \frac{\sin h\left(2\varphi \right)}{\sin \left(2\theta \right)}=\frac{\sin h\left(y\right)\cos \left(x\right)}{\sin \left(x\right)\cos h\left(y\right)}\because \left[\frac{\sin h\left(a\right)}{\cos h\left(a\right)}=\tan h\left(a\right),\frac{\cos \left(a\right)}{\sin \left(a\right)}=\cot \left(a\right)\right] \\ \Rightarrow \frac{\sin h\left(2\varphi \right)}{\sin \left(2\theta \right)}=\tan h\left(y\right)\cot \left(x\right) \\ \Rightarrow \frac{\sin h\left(2\varphi \right)}{\tan h\left(y\right)}=\sin \left(2\theta \right)\cot \left(x\right) \\ \Rightarrow \sin h\left(2\varphi \right)\cot h\left(y\right)=\sin \left(2\theta \right)\cot \left(x\right)\because \left[\frac{1}{\tan h\left(a\right)}=\cot h\left(a\right)\right] \end{gathered}$$

Hence, option (C) is correct.