If θ=tan−1d1+a1a2+tan−1d1+a2a3+⋯+tan−1d1+an&#x

Question

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}}

, where

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

are in A.P. with common difference d, then

t a n θ =

Options:

A . (n−1)d1+a1an

B . (n−1)da1+an

C . an−a1an+a1

D . nd1+a1an

Answer: Option A
:
A

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

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

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

= t a n^{- 1} \frac{a_{n} - a_{1}}{1 + a_{1} a_{n}} = t a n^{- 1} \frac{(n - 1) d}{1 + a_{1} a_{n}}

∴ t a n θ = \frac{(n - 1) d}{1 + a_{1} a_{n}}

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