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

detailed solution

Correct option is C

Since,θ1,θ2,θ3,…,θn are in AP⇒ θ2−θ1=θ3−θ2=…=θn−θn−1=d      (i)Now, taking only first termsin⁡dsec⁡θ1sec⁡θ2=sin⁡dcos⁡θ1cos⁡θ2=sin⁡θ2−θ1cos⁡θ1cos⁡θ2=sin⁡θ2cos⁡θ1−cos⁡θ2sin⁡θ1cos⁡θ1cos⁡θ2=sin⁡θ2cos⁡θ1cos⁡θ1cos⁡θ2−cos⁡θ2sin⁡θ1cos⁡θ1cos⁡θ2=tan⁡θ2−tan⁡θ1Similarly, we can solve other terms which will betan⁡θ3−tan⁡θ2,tan⁡θ4−tan⁡θ3,…∴sin⁡dsec⁡θ1sec⁡θ2+sec⁡θ2sec⁡θ3+…+sec⁡θn−1sec⁡θn=tan⁡θ2−tan⁡θ1+tan⁡θ3−tan⁡θ2 +…+tan⁡θn−tan⁡θn−1 =−tan⁡θ1+tan⁡θn=tan⁡θn−tan⁡θ1