If $$x=\frac{\sqrt3}{2}$$​, then find the value of $$\sqrt{1+x}+\sqrt{1-x}$$​.

Correct option is D

Given:

x = $$\frac{\sqrt3}2$$​​

Then, $$\sqrt{1+x}+\sqrt{1-x}$$ = ?

Formula Used:

(a + b)² = a² + b² + 2ab

(a + b)(a - b) = a² - b²​

Solution:

$$(\sqrt{1+x}+\sqrt{1-x})^2$$

= 1 + x + 1 - x + 2$$\sqrt{1-x^2}$$ 

= 2 + 2$$\sqrt{1-\frac{3}{4}}$$

= 2 + 2$$\times \sqrt\frac{1}{4}$$

= 2 + 2$$\times \frac 12$$ = 2 + 1 =3

Now,

​$$\sqrt{1+x}+\sqrt{1-x}$$ = $$\sqrt 3$$​​