Let t = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\sqrt{\mathit{x}}$
So $\sqrt{\mathit{x}}$ = tan t
i.e., $\mathit{t}\mathit{a}{\mathit{n}}^2$ t = x
On substituting x in the R.H.S. of equation $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\sqrt{\mathit{x}}$ = $\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{1-\mathit{x}}{1+\mathit{x}}\right)$
we get $\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{1-\mathit{x}}{1+\mathit{x}}\right)$ = $\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{1-\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{t}}{1+\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{t}}\right)$
Now, using the formula 2θ = $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{1-\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{t}}{1+\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{t}}\right)$ we have
$\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{1-\mathit{x}}{1+\mathit{x}}\right)$ = $\frac{1}{2}\mathit{c}\mathit{o}{\mathit{s}}^{-1}$ *cos (2t))
= t = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\sqrt{\mathit{x}}$ = L.H.S.
Hence Proved.
OR
Let a be in I quadrant such that
$\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{12}{13}\right)$ = a
So cos a = $\frac{12}{13}$
⇒ sin a = $\sqrt{1-{\left(\frac{12}{13}\right)}^2}$
= $\sqrt{1-\frac{144}{169}}$
= $\sqrt{\frac{169-144}{169}}$
= $\sqrt{\frac{25}{169}}$ = $\frac{5}{13}$
And tan a = $\frac{5}{12}$
So, a = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{5}{12}\right)$ ... (1)
Again b ∊ I quadrant such that $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{1}{3}\right)$ = b
So, sin b = $\frac{3}{5}$
⇒ cos b = $\sqrt{1-{\left(\frac{3}{5}\right)}^2}$
= $\sqrt{1-\frac{9}{25}}$
= $\sqrt{\frac{16}{25}}$ = $\frac{4}{5}$
And tan b = $\frac{3}{4}$
So, b = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{3}{4}\right)$ ... (2)
Now, let $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{56}{65}\right)$ = c where c is in I quadrant
So, sin c = $\frac{56}{65}$
⇒ cos c = $\sqrt{1-{\left(\frac{56}{65}\right)}^2}$
= $\sqrt{1-\frac{3126}{4225}}$
= $\sqrt{\frac{4225-3126}{4225}}$ = $\sqrt{\frac{1089}{4225}}$ = $\frac{33}{65}$
Ans, tan c = $\frac{56}{33}$
So c = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{56}{33}\right)$
⇒ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{56}{33}\right)$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{56}{33}\right)$ ... (3)
Now, we need t prove $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{12}{13}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$ = $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{56}{65}\right)$
Consider a + b
= $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{12}{13}\right)+\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{5}{12}\right)+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{3}{4}\right)$
[$\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{12}{13}\right)=\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{5}{12}\right)$ and $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)=\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{3}{4}\right)$]
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\frac{5}{12}+\frac{3}{4}}{1-\left(\frac{5}{12}\times \frac{3}{4}\right)}\right)$ [Using $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{y}$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\mathit{x}+\mathit{y}}{1-\mathit{x}\mathit{y}}\right)$]
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{20+36}{48-15}\right)$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{56}{33}\right)$
= c = $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{56}{65}\right)$ [Using,eq(3)]
Hence Proved.