A spherical balloon subtends an angle $2\alpha$ at a man's eye and the elevation of its centre is $\beta .$ If $\theta$ is the elevation of the highest point of the balloon at A then tan $\theta$ is equal to 

Let $r$ be the radius of the sphere with centre $O$ 

$\lfloor \underline{OPA}=\alpha ,\overline{)\underline{QPQ}}=\beta ,\overline{)\underline{RPQ}}=\theta$

$OP=r \mathrm{cosec} \alpha$ (From $∆ OAP$)

$OQ=OP\sin \beta =r\mathrm{cosec}\alpha \sin \beta$

$PQ=OP\cos \beta =r\mathrm{cosec}\alpha \cos \beta$

$\tan \theta =\frac{QR}{PQ}=\frac{r+r\mathrm{cosec}\alpha \sin \beta }{r\mathrm{cosec}\alpha \cos \beta }=\frac{\sin \alpha +\sin \beta }{\cos \beta }$