Question

$\frac{\cot 2\theta -\cot 3\theta }{\cot \theta }$ is equal to

• A. $\frac{z-y}{x}$
• B. $\frac{y-x}{z}$
• C. $\frac{z-x}{y}$
• D. none of these

Answer 1

The correct option is B $\frac{y-x}{z}$

In $△ABD\cot \theta =\frac{BD}{AB}\Rightarrow BD=z\cot 2\theta$ ...(1)
In $△ABE\cot 3\theta =\frac{BE}{AB}\Rightarrow BE=z\cot 3\theta$ ...(2)
Using (1) and (2), we get
$\frac{\cot 2\theta -\cot 3\theta }{\cot \theta }=\frac{z\cot 2\theta -z\cot 3\theta }{z\cot \theta }=\frac{BE-BD}{z\cot \theta }$
Now,
In $△MEC\cot \theta =\frac{CE}{ME}\Rightarrow CE=y\cot \theta$ ...(3)
And in $△LDC\cot \theta =\frac{CD}{DL}\Rightarrow CD=x\cot \theta$ ...(4)
Using (3) and (4)
$\frac{\cot 2\theta -\cot 3\theta }{\cot \theta }=\frac{DE}{z\cot \theta }=\frac{CE-CD}{z\cot \theta }=\frac{y\cot \theta -x\cot \theta }{z\cot \theta }=\frac{y-x}{z}$