Question #232149

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


1
Expert's answer
2021-09-02T07:25:24-0400
x2(tan2θ+cos2θ)2xytanθ+y2sin2θ=0x ^2(\tan^2θ+\cos^2θ)−2xy\tanθ+y^2\sin^2θ=0

y2sin2θ2xytanθ+x2(tan2θ+cos2θ)=0y^2\sin^2θ−2xy\tanθ+x ^2(\tan^2θ+\cos^2θ)=0

y=xtanθ±(xtanθ)2sin2θ(x2(tan2θ+cos2θ))sin2θy=\dfrac{-x\tanθ\pm\sqrt{(x\tanθ)^2-\sin^2θ(x ^2(\tan^2θ+\cos^2θ))}}{\sin^2θ}

=xtanθ±xcosθtan2θsin2θsin2θ=\dfrac{-x\tanθ\pm x\cos \theta\sqrt{\tan^2θ-\sin^2θ}}{\sin^2θ}

=xtanθ±xsin2θsin2θ=\dfrac{-x\tanθ\pm x\sin^2 \theta}{\sin^2θ}

y1=(tanθsin2θ+1)x=(tanA)xy_1=(\dfrac{-\tan\theta}{\sin^2θ}+1)x=(\tan A)x

y2=(tanθsin2θ1)x=(tanB)xy_2=(\dfrac{-\tan\theta}{\sin^2θ}-1)x=(\tan B)x

Then


tanAtanB=(tanθsin2θ+1)(tanθsin2θ1)\tan A-\tan B=(\dfrac{-\tan\theta}{\sin^2θ}+1)-(\dfrac{-\tan\theta}{\sin^2θ}-1)

tanAtanB=2\tan A-\tan B=2


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