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Mathematics Question on Sequence and series

If a1,a2,.....,an{{a}_{1}},{{a}_{2}},.....,{{a}_{n}} are in AP with common difference d≠0,d\ne 0, then (sin⁡d)(\sin d) [sec⁡a1sec⁡a2+[\sec {{a}_{1}}\sec {{a}_{2}}+ sec⁡a2sec⁡a3+...+sec⁡an−1sec⁡an]\sec {{a}_{2}}\sec {{a}_{3}}+...+\sec {{a}_{n-1}}\sec {{a}_{n}}] is equal to

A

cot⁡an−cot⁡a1\cot {{a}_{n}}-\cot {{a}_{1}}

B

cot⁡a1−cot⁡an\cot {{a}_{1}}-\cot {{a}_{n}}

C

tan⁡an−tan⁡a1\tan {{a}_{n}}-\tan {{a}_{1}}

D

tan⁡an−tan⁡an−1\tan {{a}_{n}}-\tan {{a}_{n-1}}

Answer

tan⁡an−tan⁡a1\tan {{a}_{n}}-\tan {{a}_{1}}

Explanation

Solution

Given, d=a2−a1=a3−a2=a4−a3=.....d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}=.....
=an−an−1={{a}_{n}}-{{a}_{n-1}}
∴\therefore (sin⁡d)[sec⁡a1sec⁡a2+sec⁡a2sec⁡a3+...(\sin d)[\sec {{a}_{1}}\sec {{a}_{2}}+\sec {{a}_{2}}\sec {{a}_{3}}+... +sec⁡an−1sec⁡an]+\sec {{a}_{n-1}}\sec {{a}_{n}}]
=sin⁡dcos⁡a1cos⁡a2+sin⁡dcos⁡a2cos⁡a3+....=\frac{\sin d}{\cos {{a}_{1}}\cos {{a}_{2}}}+\frac{\sin d}{\cos {{a}_{2}}\cos {{a}_{3}}}+.... +sin⁡dcos⁡an−1cos⁡an+\frac{\sin d}{\cos {{a}_{n-1}}\cos {{a}_{n}}}
=sin⁡(a2−a1)cos⁡a1cos⁡a2+sin⁡(a3−a2)cos⁡a2cos⁡a3+....=\frac{\sin ({{a}_{2}}-{{a}_{1}})}{\cos {{a}_{1}}\cos {{a}_{2}}}+\frac{\sin ({{a}_{3}}-{{a}_{2}})}{\cos {{a}_{2}}\cos {{a}_{3}}}+.... +sin⁡(an−an−1)cos⁡an−1cos⁡an+\frac{\sin ({{a}_{n}}-{{a}_{n-1}})}{\cos {{a}_{n-1}}\cos {{a}_{n}}}
=tan⁡a2−tan⁡a1+tan⁡a3−tan⁡a2+....=\tan {{a}_{2}}-\tan {{a}_{1}}+\tan {{a}_{3}}-\tan {{a}_{2}}+.... +tan⁡an−tan⁡an−1+\tan {{a}_{n}}-\tan {{a}_{n-1}}
=tan⁡an−tan⁡a1=\tan {{a}_{n}}-\tan {{a}_{1}}