Question
$\text{If}\ \text{y}\sin\phi=\text{x}\sin(2\theta+\phi),$ prove that $(\text{x+y})\cot(\theta+\phi)=(\text{y}-\text{x})\cot\theta$
✓
Answer
We have,
$\text{y}\sin\phi=\text{x}\sin(2\theta+\phi)$
$\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}...(\text{i})$
Now,
$\frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}+1=\frac{\text{x}}{\text{y}}+1$
$\Rightarrow\ \frac{\sin\phi+\sin(2\theta+\phi)}{\sin(2\theta+\phi)}=\frac{\text{x+y}}{\text{y}}...(\text{ii})$
Again,
$\frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}$ [By equation (i)]
$\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}-1=\frac{\text{x}}{\text{y}}-1$
$\Rightarrow\ \frac{\sin\phi-\sin(2\theta+\phi)}{\sin(2\theta+\phi)}=\frac{\text{x}-\text{y}}{\text{y}}...(\text{iii})$
Dividing equation (ii) by equation (iii), we get
$\frac{\sin\phi+\sin(2\theta+\phi)}{\sin\phi-\sin(2\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{2\sin\big(\frac{\phi+2\theta+\phi}{2}\big)\cos\big(\frac{\phi-2\theta-\phi}{2}\big)}{2\sin\big(\frac{\phi-2\theta-\phi}{2}\big)\cos\big(\frac{\phi+2\theta+\phi}{2}\big)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{\sin(\theta+\phi)\cos(\theta-\phi)}{\sin(-\theta)\cos(\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{\sin(\theta+\phi)\cos(\theta)}{\cos(\theta+\phi)[-\sin(\theta)]}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{-\cot(\theta)}{\cot(\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ -(\text{x}-\text{y})\cot\theta=(\text{x+y})\cot(\theta+\phi)$
$\Rightarrow\ (\text{y}-\text{x})\cot\theta=(\text{x+y})\cot(\theta+\phi)$
$\Rightarrow\ (\text{x}+\text{y})\cot(\theta+\phi)=(\text{y}-\text{x})\cot\theta$ Hence proved.
$\text{y}\sin\phi=\text{x}\sin(2\theta+\phi)$
$\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}...(\text{i})$
Now,
$\frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}+1=\frac{\text{x}}{\text{y}}+1$
$\Rightarrow\ \frac{\sin\phi+\sin(2\theta+\phi)}{\sin(2\theta+\phi)}=\frac{\text{x+y}}{\text{y}}...(\text{ii})$
Again,
$\frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}$ [By equation (i)]
$\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}-1=\frac{\text{x}}{\text{y}}-1$
$\Rightarrow\ \frac{\sin\phi-\sin(2\theta+\phi)}{\sin(2\theta+\phi)}=\frac{\text{x}-\text{y}}{\text{y}}...(\text{iii})$
Dividing equation (ii) by equation (iii), we get
$\frac{\sin\phi+\sin(2\theta+\phi)}{\sin\phi-\sin(2\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{2\sin\big(\frac{\phi+2\theta+\phi}{2}\big)\cos\big(\frac{\phi-2\theta-\phi}{2}\big)}{2\sin\big(\frac{\phi-2\theta-\phi}{2}\big)\cos\big(\frac{\phi+2\theta+\phi}{2}\big)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{\sin(\theta+\phi)\cos(\theta-\phi)}{\sin(-\theta)\cos(\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{\sin(\theta+\phi)\cos(\theta)}{\cos(\theta+\phi)[-\sin(\theta)]}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ \frac{-\cot(\theta)}{\cot(\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$
$\Rightarrow\ -(\text{x}-\text{y})\cot\theta=(\text{x+y})\cot(\theta+\phi)$
$\Rightarrow\ (\text{y}-\text{x})\cot\theta=(\text{x+y})\cot(\theta+\phi)$
$\Rightarrow\ (\text{x}+\text{y})\cot(\theta+\phi)=(\text{y}-\text{x})\cot\theta$ Hence proved.
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