If a1, a2, ..., an are in A.P with common difference d≠0, then (sin d) [sec a1 sec a2+sec a2 sec a3+ ... +sec an−1sec an] is equal to

Given, common difference =d=a2−a1=a3−a2=a4−a3= .......... =an−an−1 ∴ (sin d) [sec a1 sec a2+sec a2 sec a3+ ... +sec an−1sec an]=sindcos a1 cos a2+sindcos a2 cos a3+ ... +sindcos an−1 cos an=sin(a2−a1)cos a1 cos a2+sin(a3−a2)cos a2 cos a3+ ... +sin(an−an−1)cos an−1 cos an since sina2-a1cos a1 cosa2= sin a2 cosa1- cosa2sina1cos a1cos a2 = tan a2-tan a1 , similarly the remaining terms =tan a2−tan a1+tan a3−tan a2+ ........ +tan an−tan an−1=tan an−tan a1

If a1, a2, ..., an are in A.P with common difference d≠0, then (sin d) [sec a1 sec a2+sec a2 sec a3+ ... +sec an−1sec an] is equal to

Given, common difference =d=a2−a1=a3−a2=a4−a3= .......... =an−an−1 ∴ (sin d) [sec a1 sec a2+sec a2 sec a3+ ... +sec an−1sec an]=sindcos a1 cos a2+sindcos a2 cos a3+ ... +sindcos an−1 cos an=sin(a2−a1)cos a1 cos a2+sin(a3−a2)cos a2 cos a3+ ... +sin(an−an−1)cos an−1 cos an since sina2-a1cos a1 cosa2= sin a2 cosa1- cosa2sina1cos a1cos a2 = tan a2-tan a1 , similarly the remaining terms =tan a2−tan a1+tan a3−tan a2+ ........ +tan an−tan an−1=tan an−tan a1

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If $a_{1}, a_{2}, ..., a_{n}$ are in A.P with common difference $d \neq 0,$ then $(\sin d) [\sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + ... + \sec a_{n - 1} \sec a_{n}]$ is equal to

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$\cot a_{n} - \cot a_{1}$

$\cot a_{1} - \cot a_{n}$

$\tan a_{n} - \tan a_{1}$

$\tan a_{n} - \tan a_{n - 1}$

answer is C.

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Detailed Solution

Given, $c o m m o n d i f f e r e n c e = d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = .......... = a_{n} - a_{n - 1}$

$∴$ $(\sin d) [\sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + ... + \sec a_{n - 1} \sec a_{n}]$

$= \frac{\sin d}{\cos a_{1} \cos a_{2}} + \frac{\sin d}{\cos a_{2} \cos a_{3}} + ... + \frac{\sin d}{\cos a_{n - 1} \cos a_{n}}$

$= \frac{\sin (a_{2} - a_{1})}{\cos a_{1} \cos a_{2}} + \frac{\sin (a_{3} - a_{2})}{\cos a_{2} \cos a_{3}} + ... + \frac{\sin (a_{n} - a_{n - 1})}{\cos a_{n - 1} \cos a_{n}} (\sin c e \frac{\sin (a_{2} - a_{1})}{\cos a_{1} \cos a_{2}} = \frac{\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1}}{\cos a_{1} \cos_{} a_{2}} = \tan a_{2} - \tan a_{1}, s i m i l a r l y t h e r e m a i n i n g t e r m s)$