If an- a2- a3- - are in A-P- with common difference d- then the sum of the series sin d- a1- sec a2- sec a2 sec a3- sec an-1 sec -an- i

The correct option is D tan a_n tan a_1 We have, sin d[sec a_1 sec a_2 + sec a_2 sec a_3 + …… + sec a_n 1 sec a_n] =sin d/cos a_1 cos a_2 + sin d/cos a_2 c ...

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Byju's Answer

Standard XII

Mathematics

Common Difference

If an, a2, a3...

Question

If $a_{n}, a_{2}, a_{3}, \dots \dots$ are in A.P., with common difference d, then the sum of the series $s i n d [s e c a_{1} s e c a_{2} + s e c a_{2} s e c a_{3} + \dots \dots + s e c a_{n - 1} s e c a_{n}]$ , is

$s e c a_{1} - s e c a_{n}$

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$c o s e c a_{1} - c o s e c a_{n}$

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$c o t a_{1} - c o t a_{n}$

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$t a n a_{n} - t a n a_{1}$

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Solution

The correct option is D
$t a n a_{n} - t a n a_{1}$

We have,

$s i n d [s e c a_{1} s e c a_{2} + s e c a_{2} s e c a_{3} + \dots \dots + s e c a_{n} - 1 s e c a_{n}]$

$= \frac{s i n d}{c o s a_{1} c o s a_{2}} + \frac{s i n d}{c o s a_{2} c o s a_{3}} + \dots \dots + \frac{s i n d}{c o s a_{n - 1} c o s a_{n}}$

$= \frac{s i n (a_{2} - a_{1})}{c o s a_{1} c o s a_{2}} + \frac{s i n (a_{3} - a_{2})}{c o s a_{2} c o s a_{3}} + \dots \dots + \frac{s i n (a_{n} - a_{n - 1})}{c o s a_{n - 1} c o s a_{n}}$

$= \frac{s i n a_{2} c o s a_{1} - c o s a_{2} s i n a_{1}}{c o s a_{1} c o s a_{2}} + \dots \dots + \frac{s i n a_{3} c o s a_{2} - c o s a_{3} s i n a_{2}}{c o s a_{1} c o s a_{2}} \frac{s i n a_{2} c o s a_{1} - c o s a_{2} s i n a_{1}}{c o s a_{1} c o s a_{2}}$

$= (t a n a_{1} - t a n a_{2}) + (t a n a_{2} - t a n a_{3}) + \dots \dots + (t a n a_{n - 1} - t a n a_{n})$

$= t a n a_{1} - t a n a_{n}$

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Similar questions

Q. 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

If $a_{1}, a_{2}, a_{3}, \dots \dots a_{n}$ are in A.P., with common difference d, then the sum of the series $s i n d [c o s e c a_{1} c o s e c a_{2} + c o s e c a_{1} c o s e c a_{3} + \dots \dots + c o s e c a_{n} - 1 c o s e c a_{n}]$ is