Question

If \(a_1,a_2,a_3,\ldots ,a_n\) are in arithmetic progression with common difference d, then \(tan[tan^{-1}(\frac{d}{1+a_1a_2})+tan^{-1}(\frac{d}{1+a_2a_3})+\ldots +tan^{-1}(\frac{d}{1+a_{n-1}a_n})] =\) ____

• A. \(\frac{a_1-a_n}{1+a_1a_n}\)
• B. \(\frac{a_n-a_1}{1-a_1a_n}\)
• C. \(\frac{a_n-a_1}{1+a_1a_n}\)
• D. \(\frac{a_1+a_n}{1+a_1a_n}\)

Hint:

Write each term as tan inverse of a_{k+1} minus tan inverse of a_k.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For an AP with common difference \(d\), each term of the form \(\tan^{-1}\frac{d}{1+a_ka_{k+1}}\) can be written as a difference of two inverse tangents.

Step 2: Key Formula:
\[ \tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy} \]
Put \(x = a_{k+1}\) and \(y = a_k\). Since \(a_{k+1} - a_k = d\), we get
\[ \tan^{-1}\frac{d}{1+a_ka_{k+1}} = \tan^{-1}a_{k+1} - \tan^{-1}a_k \]

Step 3: Detailed Explanation:
The sum telescopes:
\[ (\tan^{-1}a_2 - \tan^{-1}a_1) + (\tan^{-1}a_3 - \tan^{-1}a_2) + \dots + (\tan^{-1}a_n - \tan^{-1}a_{n-1}) = \tan^{-1}a_n - \tan^{-1}a_1 \]
Taking tangent of both sides:
\[ \tan(\tan^{-1}a_n - \tan^{-1}a_1) = \frac{a_n - a_1}{1 + a_1a_n} \]

Step 4: Why the other options are wrong.
Options (A) and (D) have the wrong numerator (\(a_1 - a_n\) has the wrong sign, and \(a_1 + a_n\) is not a difference). Option (B) has \(1 - a_1a_n\) in the denominator, which belongs to the tangent of a sum, not of a difference.

Final Answer:
The value is \(\dfrac{a_n - a_1}{1 + a_1a_n}\), option (C). \[ \boxed{\frac{a_n-a_1}{1+a_1a_n}} \]