α,β are the eccentric angles of the extremities of a focal chord of the ellipse x2/16+y2/9=1 then tan⁡(α/2)tan⁡(β/2)=

The eccentricity e=1−916=74Let P(4cos⁡α,3sin⁡α) and Q(4cos⁡β,3sin⁡β) be a focalchord of the ellipse passing through the focus at (7,0)then,⇒3sin⁡β4cos⁡β−7=3sin⁡α4cos⁡α−7⇒sin⁡(α−β)sin⁡α−sin⁡β=74⇒cos⁡[(α−β)/2]cos⁡[(α+β)/2]=74⇒tan⁡α2tan⁡β2=7−47+4=23−87−9