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: Rewrite one term:
The tangent subtraction rule says $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$. With $\tan A = a_{k+1}$ and $\tan B = a_k$, the fraction is $\frac{d}{1 + a_ka_{k+1}}$.

Step 2: Telescope:
Every middle term cancels, leaving $\tan^{-1}a_n - \tan^{-1}a_1$.

Step 3: Take the tangent:
$\tan$ of that difference is $\frac{a_n - a_1}{1 + a_1a_n}$.

Step 4: Test with numbers:
Take $a = 1, 2, 3$, $d = 1$: terms are $\tan^{-1}\frac13 + \tan^{-1}\frac17$, with tangent $\frac{1/3+1/7}{1-1/21} = \frac{10/21}{20/21} = \frac12$. Formula: $\frac{3-1}{1+3} = \frac12$. It matches.

Final Answer:
The expression equals $(a_n - a_1)/(1 + a_1a_n)$, option (C). \[ \boxed{\frac{a_n-a_1}{1+a_1a_n}} \]