If ${\mathrm{\theta }}_1,{\mathrm{\theta }}_2,{\mathrm{\theta }}_3,\dots ,{\mathrm{\theta }}_{\mathrm{n}}$ are in AP, whose common difference is d, then $\sin \mathrm{d}\left(\sec {\mathrm{\theta }}_1\sec {\mathrm{\theta }}_2+\sec {\mathrm{\theta }}_2\sec {\mathrm{\theta }}_3\right.$$\left.+\dots +\sec {\mathrm{\theta }}_{\mathrm{n}-1}\sec {\mathrm{\theta }}_{\mathrm{n}}\right)$ is equal to

Since,${\mathrm{\theta }}_1,{\mathrm{\theta }}_2,{\mathrm{\theta }}_3,\dots ,{\mathrm{\theta }}_{\mathrm{n}}$ are in AP

$\Rightarrow {\mathrm{\theta }}_2-{\mathrm{\theta }}_1={\mathrm{\theta }}_3-{\mathrm{\theta }}_2=\dots ={\mathrm{\theta }}_{\mathrm{n}}-{\mathrm{\theta }}_{\mathrm{n}-1}=\mathrm{d} \left(\mathrm{i}\right)$

Now, taking only first term

$\begin{array}{l}\sin \mathrm{dsec}{\mathrm{\theta }}_1\sec {\mathrm{\theta }}_2=\frac{\sin \mathrm{d}}{\cos {\mathrm{\theta }}_1\cos {\mathrm{\theta }}_2}=\frac{\sin \left({\mathrm{\theta }}_2-{\mathrm{\theta }}_1\right)}{\cos {\mathrm{\theta }}_1\cos {\mathrm{\theta }}_2}\\ =\frac{\sin {\mathrm{\theta }}_2\cos {\mathrm{\theta }}_1-\cos {\mathrm{\theta }}_2\sin {\mathrm{\theta }}_1}{\cos {\mathrm{\theta }}_1\cos {\mathrm{\theta }}_2}\\ =\frac{\sin {\mathrm{\theta }}_2\cos {\mathrm{\theta }}_1}{\cos {\mathrm{\theta }}_1\cos {\mathrm{\theta }}_2}-\frac{\cos {\mathrm{\theta }}_2\sin {\mathrm{\theta }}_1}{\cos {\mathrm{\theta }}_1\cos {\mathrm{\theta }}_2}\\ =\tan {\mathrm{\theta }}_2-\tan {\mathrm{\theta }}_1\end{array}$

Similarly, we can solve other terms which will be

$\begin{array}{l}\tan {\mathrm{\theta }}_3-\tan {\mathrm{\theta }}_2,\tan {\mathrm{\theta }}_4-\tan {\mathrm{\theta }}_3,\dots \\ \begin{array}{ll}\therefore \sin \mathrm{d}\left(\sec {\mathrm{\theta }}_1\sec {\mathrm{\theta }}_2+\sec {\mathrm{\theta }}_2\sec {\mathrm{\theta }}_3\right.& \left.+\dots +\sec {\mathrm{\theta }}_{\mathrm{n}-1}\sec {\mathrm{\theta }}_{\mathrm{n}}\right)\\ =\tan {\mathrm{\theta }}_2-\tan {\mathrm{\theta }}_1+\tan {\mathrm{\theta }}_3-\tan {\mathrm{\theta }}_2& \\ +\dots +\tan {\mathrm{\theta }}_{\mathrm{n}}-\tan {\mathrm{\theta }}_{\mathrm{n}-1}& \\ =-\tan {\mathrm{\theta }}_1+\tan {\mathrm{\theta }}_{\mathrm{n}}=\tan {\mathrm{\theta }}_{\mathrm{n}}-\tan {\mathrm{\theta }}_1& \end{array}\end{array}$