If tan$$\ (\frac{\theta}{2})tan(\frac{2\theta}{5})\ $$= 1, then what is the value (in degrees) of $$\theta$$?

Given : $$tan(\frac{\theta}{2})tan(\frac{2\theta}{5})=1$$ --------------(i)

Now, we know that $$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$$

=> $$tan(\frac{\theta}{2}+\frac{2\theta}{5})=[tan(\frac{\theta}{2})+tan(\frac{2\theta}{5})]\div[1-tan(\frac{\theta}{2})tan(\frac{2\theta}{5})]$$

Substituting value from equation (i), we get :

=> $$tan(\frac{\theta}{2}+\frac{2\theta}{5})=\frac{tan(\frac{\theta}{2})+tan(\frac{2\theta}{5})}{0}$$

=> $$tan(\frac{\theta}{2}+\frac{2\theta}{5})=tan(90^\circ)$$

=> $$\frac{\theta}{2}+\frac{2\theta}{5}=90$$

=> $$\frac{5\theta+4\theta}{10}=90$$

=> $$\theta=90\times\frac{10}{9}$$

=> $$\theta=100^\circ$$

=> Ans - (C)