A) \[\cot {{a}_{n}}-\cot {{a}_{1}}\]
B) \[\cot {{a}_{1}}-\cot {{a}_{n}}\]
C) \[\tan {{a}_{n}}-\tan {{a}_{1}}\]
D) \[\tan {{a}_{n}}-\tan {{a}_{n-1}}\]
E) \[\tan {{a}_{1}}-\tan {{a}_{n}}\]
Correct Answer: C
Solution :
Given, \[d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}=.....\] \[={{a}_{n}}-{{a}_{n-1}}\] \[\therefore \]\[(\sin d)[\sec {{a}_{1}}\sec {{a}_{2}}+\sec {{a}_{2}}\sec {{a}_{3}}+...\] \[+\sec {{a}_{n-1}}\sec {{a}_{n}}]\] \[=\frac{\sin d}{\cos {{a}_{1}}\cos {{a}_{2}}}+\frac{\sin d}{\cos {{a}_{2}}\cos {{a}_{3}}}+....\] \[+\frac{\sin d}{\cos {{a}_{n-1}}\cos {{a}_{n}}}\] \[=\frac{\sin ({{a}_{2}}-{{a}_{1}})}{\cos {{a}_{1}}\cos {{a}_{2}}}+\frac{\sin ({{a}_{3}}-{{a}_{2}})}{\cos {{a}_{2}}\cos {{a}_{3}}}+....\] \[+\frac{\sin ({{a}_{n}}-{{a}_{n-1}})}{\cos {{a}_{n-1}}\cos {{a}_{n}}}\] \[=\tan {{a}_{2}}-\tan {{a}_{1}}+\tan {{a}_{3}}-\tan {{a}_{2}}+....\] \[+\tan {{a}_{n}}-\tan {{a}_{n-1}}\] \[=\tan {{a}_{n}}-\tan {{a}_{1}}\]
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