$\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}\right)$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ $\left[\frac{\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }}{2}-\mathit{x}\right)}{1+\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\pi }}{2}-\mathit{x}\right)}\right]$
= [Since sin θ = 2 sin (θ/2) cos (θ/2) and 1 + cos θ = 2 $\mathit{c}\mathit{o}{\mathit{s}}^2$ (θ/2)]
$\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ $\left[\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}-\frac{\mathit{x}}{2}\right)\right]$ = $\left(\frac{\mathit{\pi }}{4}-\frac{\mathit{x}}{2}\right)$ (proved)
OR
Let $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{8}{17}$ = x
Then , sin x = $\frac{8}{17}$ ; cos x = $\sqrt{1-{\mathit{x}}^2}$
⇒ cos x = $\sqrt{1-{\left(\frac{8}{17}\right)}^2}$
⇒ cos x = $\sqrt{1-{\left(\frac{8}{17}\right)}^2}$
⇒ cos x = $\sqrt{\frac{225}{289}}$
⇒ cos x = $\frac{15}{17}$
∴ tan x = $\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{x}}{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}$
⇒ tan x = $\frac{\frac{8}{17}}{\frac{15}{17}}$
⇒ tan x = $\frac{8}{15}$
⇒ x = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{8}{15}\right)$ ... (1)
Let $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{3}{5}$ = y ... (2)
Then, sin y = $\frac{3}{5}$ ; cos y = $\sqrt{1-{\mathit{y}}^2}$
⇒ cos y = $\sqrt{1-{\left(\frac{3}{5}\right)}^2}$
⇒ cos y = $\sqrt{\frac{16}{25}}$
⇒ cos y = $\frac{4}{5}$
∴ tan y = $\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{y}}{\mathit{c}\mathit{o}\mathit{s}\mathit{y}}$
⇒ tan y = $\frac{\frac{3}{5}}{\frac{4}{5}}$
⇒ tan y = $\frac{3}{4}$
⇒ y = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{3}{4}\right)$ ... (3)
From equations (2) and (3), we have,
$\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{3}{4}\right)$
Now consider $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{8}{17}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$
From equations (1) and (3), we have, $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{8}{17}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{5}{15}\right)+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{3}{4}\right)$
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}\right)$ [Since $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{y}$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\frac{\mathit{x}+\mathit{y}}{\mathit{x}-\mathit{y}}$]
= $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{32+45}{60-24}\right)$
$\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{8}{17}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{77}{36}\right)$ ... (4)
Now, we have:
Let $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{77}{36}\right)$ = z
Then tan z = $\frac{77}{36}$
⇒ sec z = $\sqrt{1+{\left(\frac{77}{36}\right)}^2}$ [Since sec θ = $\sqrt{1+\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }}$]
⇒ sec z = $\sqrt{\frac{1296+5929}{1296}}$
⇒ sec z = $\sqrt{\frac{7225}{1296}}$
⇒ sec z = $\frac{85}{36}$
We know that cosz = $\frac{1}{\mathit{s}\mathit{e}\mathit{c}\mathit{z}}$
Thus, sec z = $\frac{85}{36}$ , cos z = $\frac{36}{85}$
⇒ z = $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{36}{85}\right)$
⇒ $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\frac{77}{36}\right)$ = $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{36}{85}\right)$
⇒ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{8}{17}\right)$ + $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{3}{5}\right)$ = $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\left(\frac{36}{85}\right)$ [Since from equation (4)]
Hence proved.