See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

tan  = usin tan eucos e      tan  = 1 tan e   tan  = 1 tan e       2 + 2 + 2 = 180°  +  = (90 – ) tan( + ) = cot   tan tan 1 1 tan tan tan        2 2 3 1 1 tan tan 1 e e 1 tan 1 tan e       AITS-CRT-I-PCM (Sol.)-JEE(Main)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 6  2 2 2 3 1 1 1 tan 1 tan e e e            2 2 3 1 1 1 tan 1 e e e           3 2 2 2 3 1 e tan 1 1 1 1 e e e e e                  