If $$21 \tan \theta = 20$$, then $$(1 + \sin \theta + \cos \theta) : (1 - \sin \theta + \cos \theta) = ?$$

Given that $$21 \tan \theta = 20$$

$$ \tan \theta = \dfrac{20}{21}$$

then $$(1 + \sin \theta + \cos \theta) : (1 - \sin \theta + \cos \theta)$$

$$\dfrac{(1 + \sin \theta + \cos \theta)}{ (1 - \sin \theta + \cos \theta)}$$

dividing   $$\cos \theta $$

$$\dfrac {(\sec \theta + \tan \theta + 1)}{(\sec \theta - \tan \theta + 1)} $$

we know that $$\sec^2 \theta - \tan^2 \theta = 1 $$

then $$\Rightarrow \dfrac{(\sec \theta +\tan \theta) (\sec^2 \theta - \tan^2 \theta)} {(\sec \theta - \tan \theta + 1)}$$

$$\Rightarrow (\sec \theta + \tan \theta) $$

Now $$ \tan \theta = \dfrac{20}{21} $$

we know that $$ 1 + \tan^2 \theta = \sec^2 \theta $$

$$\Rightarrow 1 + \dfrac{400}{441} = \sec^2 \theta $$ 

$$\Rightarrow \sec \theta = \dfrac{29}{21}$$

hence $$\sec \theta + \tan \theta $$

$$\Rightarrow \dfrac{29}{21}+ \dfrac{20}{21}$$ (put the value)

$$\Rightarrow \dfrac{49}{21}$$

$$\Rightarrow \dfrac{7}{3} $$ Ans