A) \[\sec {{a}_{1}}-\sec {{a}_{n}}\]
B) \[\cot {{a}_{1}}-\cot {{a}_{n}}\]
C) \[\tan {{a}_{1}}-\tan {{a}_{n}}\]
D) \[c\text{osec}\ {{a}_{1}}-\text{cosec}\ {{a}_{n}}\]
Correct Answer: B
Solution :
As given \[d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=....={{a}_{n}}-{{a}_{n-1}}\] \[\therefore \]\[\sin d\,\{co\text{sec}\ {{a}_{1}}co\text{sec}\ {{a}_{2}}+.....+\text{cosec}\ {{a}_{n-1}}\text{cosec}\ {{a}_{n}}\}\] \[=\frac{\sin ({{a}_{2}}-{{a}_{1}})}{\sin {{a}_{1}}.\ \sin {{a}_{2}}}+......+\frac{\sin ({{a}_{n}}-{{a}_{n-1}})}{\sin {{a}_{n-1}}\sin {{a}_{n}}}\] \[=(\cot {{a}_{1}}-\cot {{a}_{2}})+(\cot {{a}_{2}}-\cot {{a}_{3}})+....\]\[+(\cot {{a}_{n-1}}-\cot {{a}_{n}})\] \[=\cot {{a}_{1}}-\cot {{a}_{n}}\].
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