Prove that $\sqrt {\dfrac{{1 - \cos A}}{{1 + \cos A}}} + \sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} = 2\cos ecA$

Hint: To solve this problem we need trigonometry which helps to prove L.H.S=R.H.S. We need to know the rationalization concept as question demands.

Complete step-by-step answer:

Let us take the L.H.S part
$ \Rightarrow \sqrt {\dfrac{{1 - \cos A}}{{1 + \cos A}}} + \sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}}} $
Now on rationalizing the denominators by multiplying suitable factors, we will get
$ \Rightarrow \sqrt {\dfrac{{1 - \cos A}}{{1 + \cos A}} \times \dfrac{{1 - \cos A}}{{1 - \cos A}}} + \sqrt {\dfrac{{1 + \cos A}}{{1 - \cos A}} \times \dfrac{{1 + \cos A}}{{1 + \cos A}}} $
\[\begin{gathered}
   \Rightarrow \sqrt {\dfrac{{{{(1 - \cos A)}^2}}}{{(1 + \cos A)(1 - \cos A)}}} + \sqrt {\dfrac{{{{(1 + \cos A)}^2}}}{{(1 - \cos A)(1 + \cos A)}}} \\
   \Rightarrow \sqrt {\dfrac{{{{(1 - \cos A)}^2}}}{{({1^2} - {{\cos }^2}A)}}} + \sqrt {\dfrac{{{{(1 + \cos A)}^2}}}{{({1^2} - {{\cos }^2}A)}}} \\
\end{gathered} \]
Here in the above term of numerator square root and square get cancels and in denominator we know that $\sin A = \sqrt {1 - {{\cos }^2}A} $
On simplifying the above term we get
$ \Rightarrow \dfrac{{1 - \cos A}}{{sinA}} + \dfrac{{1 + \cos A}}{{\sin A}}$
$\begin{gathered}
   \Rightarrow \dfrac{{1 - \cos A + 1 + \cos A}}{{\sin A}} \\
   \Rightarrow \dfrac{2}{{\sin A}} \\
\end{gathered} $
We know that $\sin A = \dfrac{1}{{\cos ecA}}$ now on replacing the term we get it as
$\begin{gathered}
   \Rightarrow 2\cos ecA \\
   \Rightarrow R.H.S \\
\end{gathered} $
Hence we have proved L.H.S=R.H.S

NOTE: In this problem we have taken the L.H.S part and rationalize the denominator by using its multiplying suitable factor. And by using the basic formulas and conversion we got the R.H.S value. Hence we have proved that L.H.S=R.H.S.