Question

If \[\theta \] is the angle which the straight line joining the points \[\left( {{x}_{1}},{{y}_{1}} \right)\]and \[\left( {{x}_{2}},{{y}_{2}} \right)\]subtends at the origin prove that \[\tan \theta =\dfrac{{{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}\] and \[\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}\sqrt{{{x}_{2}}^{2}+{{y}_{2}}^{2}}}\].

Answer 1

Hint:Find the slope of the line formed by point A and B with origin because that will be the angle subtended by line AB on the origin , then use the Formula \[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]to find \[\theta \] to find the angle between those two lines.

Complete step-by-step answer:
Let A \[\left( {{x}_{1}},{{y}_{1}} \right)\] and B \[\left( {{x}_{2}},{{y}_{2}} \right)\] be the given points.
Let O be the origin.
Slope of OA = \[{{m}_{1}}=\dfrac{{{y}_{1}}}{{{x}_{1}}}\]
Slope of OB = \[{{m}_{2}}=\dfrac{{{y}_{2}}}{{{x}_{2}}}\]
It is given that \[\theta \] is the angle between lines OA and OB.
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
By substituting the values of A and B we get,
\[\tan \theta =\dfrac{\dfrac{{{y}_{1}}}{{{x}_{1}}}-\dfrac{{{y}_{2}}}{{{x}_{2}}}}{1+\left( \dfrac{{{y}_{1}}}{{{x}_{1}}}\times \dfrac{{{y}_{2}}}{{{x}_{2}}} \right)}\]
By solving this we get,
\[\tan \theta =\dfrac{{{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that \[\cos \theta =\sqrt{\dfrac{1}{1+{{\tan }^{2}}\theta }}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substitute the value (1) in (2), we get
\[\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{({{x}_{2}}{{y}_{1}}-{{x}_{1}}{{y}_{2}})}^{2}}+{{\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)}^{2}}}}\]
\[\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{x}_{2}}^{2}{{y}_{1}}^{2}+{{x}_{1}}^{2}{{y}_{2}}^{2}+{{x}_{1}}^{2}{{x}_{2}}^{2}+{{y}_{1}}^{2}{{y}_{2}}^{2}}}\]
\[\cos \theta =\dfrac{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}}{\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}\sqrt{{{x}_{2}}^{2}+{{y}_{2}}^{2}}}\]
Hence proved.

Note: From the above we can conclude that after finding the slopes we get the value \[\tan \theta \] and further substituted in the trigonometric operation to get the final solution. Only variables are used so modulus doesn’t play an important role.