MHTCET 20 Apr 2026 Shift 1 Paper

Concept:Use the identity $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$ to rewrite each term, making the sum a telescoping series.Explanation:Since $a_1, a_2, \ldots, a_n$ are in arithmetic progression with common difference $d$, we have $d = a_{r+1} - a_r$.Consider a general term of the given expression:$\tan^{-1}\left(\frac{d}{1+a_r a_{r+1}}\right)=\tan^{-1}\left(\frac{a_{r+1}-a_r}{1+a_r a_{r+1}}\right)$.Using $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, we get:$\frac{a_{r+1}-a_r}{1+a_r a_{r+1}} = \tan\left(\tan^{-1}a_{r+1}-\tan^{-1}a_r\right)$.Therefore, each term simplifies to $\tan^{-1}a_{r+1}-\tan^{-1}a_r$.Substituting this into the whole sum creates a telescoping effect:$\left(\tan^{-1}a_2-\tan^{-1}a_1\right)+\left(\tan^{-1}a_3-\tan^{-1}a_2\right)+\cdots+\left(\tan^{-1}a_n-\tan^{-1}a_{n-1}\right)$.All middle terms cancel, leaving $\tan^{-1}a_n-\tan^{-1}a_1$.Taking tangent on both sides:$\tan\left(\tan^{-1}a_n-\tan^{-1}a_1\right)=\frac{a_n-a_1}{1+a_1 a_n}$.Answer:$\frac{a_n-a_1}{1+a_1 a_n}$Hence, the correct option is C.