Question

If $x=\frac{\sqrt{3}}{2}$ then find the value of :-
$\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$

Answer 1

We have,

$\begin{array}{c}x=\frac{\sqrt{3}}{2}\\ \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}\\ x=\sin 2\theta \\ \theta ={30}^0\end{array}$

$\frac{1+\sin 2\theta }{1+\sqrt{1+\sin 2\theta }}+\frac{1-\sin 2\theta }{1-\sqrt{1-\sin 2\theta }}$

$\frac{1+\sin 2\theta }{\left(1+\sin \theta +\cos \theta \right)}+\frac{1-\sin 2\theta }{1-\left(\cos \theta -\sin \theta \right)}$

$\frac{\left(\sin \theta +1-\cot \theta \right)\left(1+\sin 2\theta \right)+\left(1-\sin 2\theta \right)\left(1+\sin \theta +\cos \theta \right)}{{\left(\sin \theta +1\right)}^2-{\cos }^2\theta }$

$=\frac{2\sin \theta +2-2\cos \theta \sin \theta }{{\left(\sin \theta +1\right)}^2-{\cos }^2\theta }$

An putting $\theta ={30}^0$

$=\frac{2\times \frac{1}{2}+2-2\frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}}{\frac{9}{4}-\frac{3}{4}}$

$=\frac{3-3/2}{6/4}=\frac{3/2}{3/2}=1$