Miscellaneous

Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise

### Transcript

Question 1 Prove that tanโ1 1/5 + tanโ1 1/7 + tanโ1 1/3 + tanโ1 1/8 = ๐/4 We know that tanโ1 x + tanโ1 y = tanโ1 ((๐ฑ + ๐ฒ )/(๐ โ ๐ฑ๐ฒ)) tanโ1 ๐/๐ + tanโ1 ๐/๐ = tanโ1 (1/5 + 1/7)/(1โ 1/5 ร 1/7) = tanโ1 ((7 + 5)/(5(7)))/( (35 โ 1)/35 ) = tanโ1 (6/17) tanโ1 ๐/๐ + tanโ1 ๐/๐ = tanโ1 (1/3 + 1/8)/(1โ 1/3 ร 1/8) = tanโ1 ( (8 + 3)/(3(8)))/( (24 โ 1)/24) = tan โ 1 (11/23) Solving L.H.S tanโ1 1/5 + tanโ1 1/7 + tanโ1 1/3 + tanโ1 1/8 = ("tanโ1 " 1/5 " + tanโ1 " 1/7) + ("tanโ1 " 1/3 " + tanโ1 " 1/8) = tan-1 (๐/๐๐) + tanโ1 (๐๐/๐๐) = tanโ1 (6/17 + 11/23)/(1 โ 6/17 ร 11/23) We know that tanโ1 x + tanโ1 y = tanโ1 ((๐ + ๐ )/(๐ โ ๐๐)) Replacing x by 6/17 and y by 11/23 = tanโ1 ( (6(23) + 11(17))/(17(23)))/(1 โ 66/391) = tanโ1 ( (138 + 187)/391)/((391 โ 66)/391) = tanโ1 ( 325/391)/(325/391) = tanโ1 1 = tanโ1 ("tan " ๐/๐) = ฯ/4 = R.H.S Hence proved (As tan ๐/4 = 1)