Answer to Question #232149 in Analytic Geometry for jobaer

Question

if two straight line represented by the equation x²(tan² + cos² )-2xytan +y² sin²=0 makes angle A and B with x axis respectively ,than show that tan A - tan B

Answer 1

$$x ^2(\tan^2θ+\cos^2θ)−2xy\tanθ+y^2\sin^2θ=0$$

$$y^2\sin^2θ−2xy\tanθ+x ^2(\tan^2θ+\cos^2θ)=0$$

$$y=\dfrac{-x\tanθ\pm\sqrt{(x\tanθ)^2-\sin^2θ(x ^2(\tan^2θ+\cos^2θ))}}{\sin^2θ}$$

$$=\dfrac{-x\tanθ\pm x\cos \theta\sqrt{\tan^2θ-\sin^2θ}}{\sin^2θ}$$

$$=\dfrac{-x\tanθ\pm x\sin^2 \theta}{\sin^2θ}$$

$$y_1=(\dfrac{-\tan\theta}{\sin^2θ}+1)x=(\tan A)x$$

$$y_2=(\dfrac{-\tan\theta}{\sin^2θ}-1)x=(\tan B)x$$

Then

$$\tan A-\tan B=(\dfrac{-\tan\theta}{\sin^2θ}+1)-(\dfrac{-\tan\theta}{\sin^2θ}-1)$$

$$\tan A-\tan B=2$$