Question

If the equation of a straight line $3x+4y+12=0$ is transformed into normal form $x\cos \alpha +y\sin \alpha =p$, then the value of $\tan \alpha +\cot \alpha$ is equal to

• A. $\frac{25}{12}$
• B. $\frac{12}{25}$
• C. $\frac{7}{12}$
• D. $\frac{25}{7}$

Answer 1

The correct option is A $\frac{25}{12}$

Given : $3x+4y+12=0\Rightarrow -3x-4y=12$
$\Rightarrow \left(\frac{-3}{5}\right)x+\left(\frac{-4}{5}\right)y=\frac{12}{5}$ (normal form of line)
$\Rightarrow \cos \alpha =-\frac{3}{5},\sin \alpha =-\frac{4}{5}$
$\Rightarrow \alpha$ is in third quadrant
$\Rightarrow \tan \alpha =\frac{4}{3},\cot \alpha =\frac{3}{4}\therefore \tan \alpha +\cot \alpha =\frac{25}{12}$