If a1- a2- a3- - an are in A-P- with common difference d- then the sum of the series sin d- a1 a2- a2- sec -a3- - an-1 an- isa a1- anb cosec a1-cosec anc a1- and tan an-tan a1

(d) tan an minus; tan a1 We have: sin #160;d #160;sec #160;a1 #160;sec #160;a2+sec #160;a2 #160;sec #160;a3+....+sec #160;an 1sec #160;a ...

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Byju's Answer

Standard IX

Mathematics

Calculating Heights and Distances

If a1, a2, a3...

Question

If a₁, a₂, a₃, .... a_n are in A.P. with common difference d, then the sum of the series sin d [sec a₁ sec a₂ + sec a₂ sec a₃ + .... + sec a_n_{− 1} sec a_n], is
(a) sec a₁ − sec a_n
(b) cosec a₁ − cosec a_n
(c) cot a₁ − cot a_n
(d) tan a_n − tan a₁

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Solution

(d) tan a_n − tan a₁
We have:

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

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If a1, a2, a3, .... an are in A.P. with common difference d, then the sum of the series sin d [sec a1 sec a2 + sec a2 sec a3 + .... + sec an − 1 sec an], is (a) sec a1 − sec an (b) cosec a1 − cosec an (c) cot a1 − cot an (d) tan an − tan a1

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(a) sec a₁ − sec a_n
(b) cosec a₁ − cosec a_n
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