Question

If the lines represented by the equation \[2{x^2} - 3xy + {y^2} = 0\] make angles \[\alpha \] and \[\beta \] with x-axis. Then what is the value of \[\cot^{2}\alpha + \cot^{2}\beta \] ?
A. 0
B. \[\dfrac{3}{2}\]
C. \[\dfrac{7}{4}\]
D. \[\dfrac{5}{4}\]

Answer 1

Hint: First, simplify the given equation of the lines and find the values of \[\dfrac{y}{x}\] . Because angle made by the line is the tan inverse of the slope of the line. i.e., \[\tan\theta = m\] or \[tan\theta = \dfrac{y}{x}\]. After that, simplify the given trigonometric equation in terms of \[tan\]. In the end, substitute the values in the equation to get the required answer.

Formula used:
\[tan x = \dfrac{1}{{cot x}}\]
Angle made by a line with x-axis is: \[\tan\theta = \dfrac{y}{x}\], where \[y = mx\]

Complete step by step solution:
Given:
The line represented by the equation \[2{x^2} - 3xy + {y^2} = 0\] make angles \[\alpha \] and \[\beta \] with x-axis.
Let’s simplify the equation of lines.
\[2{x^2} - 3xy + {y^2} = 0\]
Factorize the above equation.
\[2{x^2} - 2xy - xy + {y^2} = 0\]
\[ \Rightarrow 2x\left( {x - y} \right) - y\left( {x - y} \right) = 0\]
\[ \Rightarrow \left( {x - y} \right)\left( {2x - y} \right) = 0\]
\[ \Rightarrow x - y = 0\] or \[2x - y = 0\]
\[ \Rightarrow x = y\] or \[2x = y\]
\[ \Rightarrow \dfrac{y}{x} = 1\] or \[\dfrac{y}{x} = 2\]
We know that, if \[\theta \] is the inclination of the line, then \[\tan\theta \] is the slope or the gradient of the line.
So, we get
\[\tan\alpha = 1\] and \[\tan\beta = 2\]
Now solve the given expression \[\cot^{2}\alpha + \cot^{2}\beta \].
Apply the trigonometric identity \[\cot x = \dfrac{1}{{\tan x}}\].
\[\cot^{2}\alpha +\ cot^{2}\beta = \dfrac{1}{{\tan^{2}\alpha }} + \dfrac{1}{{\tan^{2}\beta }}\]
\[ \Rightarrow \cot^{2}\alpha + \cot^{2}\beta = \dfrac{1}{{{{\left( {\tan\alpha } \right)}^2}}} + \dfrac{1}{{{{\left( {\tan\beta } \right)}^2}}}\]
Substitute the values of \[\tan\alpha \] and \[\tan\beta \] in the above equation.
\[\cot^{2}\alpha + \cot^{2}\beta = \dfrac{1}{{{{\left( 1 \right)}^2}}} + \dfrac{1}{{{{\left( 2 \right)}^2}}}\]
\[ \Rightarrow \cot^{2}\alpha + \cot^{2}\beta = \dfrac{1}{1} + \dfrac{1}{4}\]
Simplify the right-hand side.
\[\cot^{2}\alpha + \cot^{2}\beta = 1 + \dfrac{1}{4}\]
\[ \Rightarrow \cot^{2}\alpha + \cot^{2}\beta = \dfrac{5}{4}\]
Hence the correct option is D.

Note: Students often get confused about the relation between the slope and the trigonometric function \[\tan\].
If \[\theta \] is the angle of inclination of a straight line, then \[\tan\theta \] is the slope of the line. Where \[{0^{\circ}} < \theta < 180^{\circ}\].