Let a1- a2- - an be an A-P- with common difference -6 and assume a1 a2- a2 a3- an-1 an-ktan an-tan a1 find the value of k-

Given nbsp; a_1 a_2+ a_2 a_3+....+ a_n 1 a_n.=k(tan a_n tan a_1) nbsp; nbsp;....(1)Now, nbsp; a_1 a_2+ a_2 a_3+....+ a_n 1 a_n 1co ...

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Domain and Range of Basic Inverse Trigonometric Functions

Let a1, a2,...

Question

Let $a_{1}, a_{2}, . . . ., a_{n}$ be an A.P. with common difference $\frac{π}{6}$ and assume $sec a_{1} sec a_{2} + sec a_{2} sec a_{3} + . . . . + sec a_{n - 1} sec a_{n} . = k (tan a_{n} - tan a_{1})$ find the value of $k$ .

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a_{1}, a_{2}, a_{3} \dots \dots . . a_{n}

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots \dots \dots \dots \dots \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

a_{1}, a_{2}, . . ., a_{n}

d

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a_{1}, a_{2}, a_{3}, . . ., a_{n}

tan [{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + . . . + {tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] = \frac{(n - p) d}{q + a_{1} a_{n}}

p + q

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

a_{1}, a_{2}, a_{3}, \dots a_{n}

t a n θ =

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