Question

$\frac{cot\theta -cot2\theta }{cot\theta }$ is equal to

• A. $\frac{x}{z}$
• B. $\frac{y}{z}$
• C. $\frac{y}{x}$
• D. $\frac{x}{y}$

Answer 1

The correct option is A $\frac{x}{z}$

In $△ABC$
$cot\theta =\frac{BC}{AB}\Rightarrow BC=zcot\theta$ ...(1)
In $△ABD$
$cot\theta =\frac{BD}{AB}\Rightarrow BD=zcot2\theta$ ..(2)
Now in similar $△ABC$ and $△LDC$
$\frac{CD}{BC}=\frac{DL}{AB}=\frac{x}{z}$
Therefore,
$\frac{cot\theta -cot2\theta }{cot\theta }=\frac{zcot\theta -zcot2\theta }{zcot\theta }=\frac{BC-BD}{BC}=\frac{CD}{BC}=\frac{x}{z}$