1.If a1a_1a1, a2a_2a2, a3a_3a3 are in arithmetic progression and ddd is the common difference, then tan−1(d1+a1a2)+tan−1(d1+a2a3)=\tan^{-1} \left( \frac{d}{1+a_1a_2} \right) + \tan^{-1} \left( \frac{d}{1+a_2a_3} \right) =tan−1(1+a1a2d)+tan−1(1+a2a3d)=a.tan−1(2d1+a1a3)\tan^{-1} \left( \frac{2d}{1+a_1a_3} \right)tan−1(1+a1a32d)b.tan−1(d1+a1a3)\tan^{-1} \left( \frac{d}{1+a_1a_3} \right)tan−1(1+a1a3d)c.tan−1(2d1+a2a3)\tan^{-1} \left( \frac{2d}{1+a_2a_3} \right)tan−1(1+a2a32d)d.tan−1(2d1−a1a3)\tan^{-1} \left( \frac{2d}{1-a_1a_3} \right)tan−1(1−a1a32d)Login to continueOnly logged in users canattempt or see the solution.
