Question

If $\tan x-\tan y=m$ and $\cot y-\cot x=n,$ then $(\frac{1}{m}+\frac{1}{n})$ is equal to

• A. $\sin (x-y)$
• B. $\cot (x-y)$
• C. $\tan (x-y)$
• D. $\cos (x-y)$

Answer 1

The correct option is B $\cot (x-y)$

We have,
$m=\tan x-\tan y=\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}=\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}=\frac{\sin (x-y)}{\cos x\cos y}\cdots \left(i\right)\therefore \frac{1}{m}=\frac{\cos x\cos y}{\sin (x-y)}$

Again,
$n=\cot y-\cot x=\frac{\cos y}{\sin y}-\frac{\cos x}{\sin x}=\frac{\sin x\cos y-\cos x\sin y}{\sin x\sin y}=\frac{\sin (x-y)}{\sin x\sin y}\therefore \frac{1}{n}=\frac{\sin x\sin y}{\sin (x-y)}\cdots \left(ii\right)$

Now, adding equation $\left(i\right)$ and $\left(ii\right),$ we get
$\frac{1}{m}+\frac{1}{n}=\frac{\cos x\cos y}{\sin (x-y)}+\frac{\sin x\sin y}{\sin (x-y)}=\frac{\cos x\cos y+\sin x\sin y}{\sin (x-y)}=\frac{\cos (x-y)}{\sin (x-y)}\therefore \frac{1}{m}+\frac{1}{n}=\cot (x-y)$