Circled question

Dear Student,

cosθ​ = $\frac{\mathrm{cos\alpha } \mathrm{cos\beta }}{1-\mathrm{sin\alpha } \mathrm{sin\beta }}$ 
Let $\frac{\mathrm{cos\alpha } \mathrm{cos\beta }}{1-\mathrm{sin\alpha } \mathrm{sin\beta }}$ = x
Therefore,
cosθ = x
$\frac{1-{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}{1+{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}}$ = x
1 - tan²$\frac{\mathrm{\theta }}{2}$ = x + x tan²$\frac{\mathrm{\theta }}{2}$
1 - x = tan²$\frac{\mathrm{\theta }}{2}$ + x tan²$\frac{\mathrm{\theta }}{2}$
tan²$\frac{\mathrm{\theta }}{2}$ = $\frac{1-\mathrm{x}}{1+\mathrm{x}}$      ................. (1) 
Now,
x = $\frac{\mathrm{cos\alpha } \mathrm{cos\beta }}{1-\mathrm{sin\alpha } \mathrm{sin\beta }}$ 
  = $\frac{\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{\mathrm{\alpha }}{2}} }{1+{\tan }^2\frac{\mathrm{\alpha }}{2} }}\right)\left({\displaystyle \frac{1-{\tan }^2\frac{\mathrm{\beta }}{2}}{1+{\tan }^2\frac{\mathrm{\beta }}{2}}}\right)}{1-{\displaystyle \frac{2\tan {\displaystyle \frac{\mathrm{\alpha }}{2}} }{1+{\tan }^2\frac{\mathrm{\alpha }}{2}}}{\displaystyle \frac{2\tan \frac{\mathrm{\beta }}{2}}{1+{\tan }^2\frac{\mathrm{\beta }}{2}}}}$
= $\frac{\left({\displaystyle 1-{\tan }^2\frac{{\displaystyle \mathrm{\alpha }}}{{\displaystyle 2}}}\right)\left({\displaystyle 1-{\tan }^2\frac{\mathrm{\beta }}{2}}\right)}{\left(1+{\tan }^2\frac{\mathrm{\alpha }}{2}\right)\left(1+{\tan }^2\frac{\mathrm{\beta }}{2}\right)-4\tan \frac{\mathrm{\alpha }}{2} \tan \frac{\beta }{2}}$
Now, 
tan²$\frac{\mathrm{\theta }}{2}$  = $\frac{1-\frac{\left({\displaystyle 1-{\tan }^2\frac{{\displaystyle \mathrm{\alpha }}}{{\displaystyle 2}}}\right)\left({\displaystyle 1-{\tan }^2\frac{\mathrm{\beta }}{2}}\right)}{\left(1+{\tan }^2\frac{\mathrm{\alpha }}{2}\right)\left(1+{\tan }^2\frac{\mathrm{\beta }}{2}\right)-4\tan \frac{\mathrm{\alpha }}{2} \tan \frac{\beta }{2}}}{1+\frac{\left({\displaystyle 1-{\tan }^2\frac{{\displaystyle \mathrm{\alpha }}}{{\displaystyle 2}}}\right)\left({\displaystyle 1-{\tan }^2\frac{\mathrm{\beta }}{2}}\right)}{\left(1+{\tan }^2\frac{\mathrm{\alpha }}{2}\right)\left(1+{\tan }^2\frac{\mathrm{\beta }}{2}\right)-4\tan \frac{\mathrm{\alpha }}{2} \tan \frac{\beta }{2}}}$
Let tan$\frac{\alpha }{2}$ = a and tan$\frac{\beta }{2}$ = b
tan²$\frac{\mathrm{\theta }}{2}$ = $\frac{1-{\displaystyle \frac{\left(1-{\mathrm{a}}^2\right)\left(1-{\mathrm{b}}^2\right)}{\left(1+{\mathrm{a}}^2\right)\left(1+{\mathrm{b}}^2\right)-4\mathrm{ab}}}}{1+{\displaystyle \frac{\left(1-{\mathrm{a}}^2\right)\left(1-{\mathrm{b}}^2\right)}{\left(1+{\mathrm{a}}^2\right)\left(1+{\mathrm{b}}^2\right)-4\mathrm{ab}}}}$
          = $\frac{\left(1+{\mathrm{a}}^2\right)\left(1+{\mathrm{b}}^2\right)-4\mathrm{ab}-{\displaystyle \left(1-{\mathrm{a}}^2\right)\left(1-{\mathrm{b}}^2\right)}}{\left(1+{\mathrm{a}}^2\right)\left(1+{\mathrm{b}}^2\right)-4\mathrm{ab}+{\displaystyle \left(1-{\mathrm{a}}^2\right)\left(1-{\mathrm{b}}^2\right)}}$
        = $\frac{\left(1+a^2+b^2+a^2{b}^2\right)-4\mathrm{ab}-{\displaystyle \left(1-a^2-b^2+a^2{b}^2\right)}}{\left(1+a^2+b^2+a^2{b}^2\right)-4\mathrm{ab}+{\displaystyle \left(1-a^2-b^2+a^2{b}^2\right)}}$
      = ​$\frac{2a^2+2b^2-4\mathrm{ab}}{2+2a^2{b}^2-4\mathrm{ab}}$
     = $\frac{{\left(\mathrm{a}-\mathrm{b}\right)}^2}{{\left(1-\mathrm{ab}\right)}^2}$
tan$\frac{\mathrm{\theta }}{2}$ = $\frac{\left(\mathrm{a}-\mathrm{b}\right)}{\left(1-\mathrm{ab}\right)}$
         = $\frac{\tan {\displaystyle \frac{\alpha }{2}}-\tan {\displaystyle \frac{\beta }{2}}}{1-\tan \frac{\alpha }{2} \tan \frac{\beta }{2}}$
Hence proved

Regards