Tiêu đề Inverse Trigonometry Varsity Practice Sheet Solution.pdf ✅

07 wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY Inverse Trigonometric Functions and Trigonometric Equations weMZ mv‡j DU-G Avmv cÖkœvejx 1. wb‡Pi †KvbwU sin–1 2 3 Gi mgvb? [DU 23-24] cot 5 2 cot–1 5 2 tan 5 tan–1 1 5 DËi: cot–1 5 2 e ̈vL ̈v: sin–12 3 = cot–1 5 2 3 2 – 2 2 = 5 2 3 2. 2tan–1 1 3 + tan–1 1 7 = ? [DU 23-24]  4  2 1 0 DËi:  4 e ̈vL ̈v: 2tan–1 1 3 + tan–1 1 7 = tan–1 2 3 1 – 1 9 + tan–1 1 7 = tan–1 3 4 + tan–1 1 7 = tan–1 3 4 + 1 7 1 – 3 4  1 7 = tan–1       25 28 25 28 =  4 3. tan + cot = 2cosec, 0   <  2 n‡j,  Gi gvb KZ? [DU 22-23]  4 5 3  6  3 DËi:  3 e ̈vL ̈v: tan + cot = 2cosec  sin cos + cos sin = 2 sin  1 sincos = 2 sin  sin = 2sincos  sin– 2cos nq, sin = 0   = nn  Z A_ev, 1 – 2cos = 0  cos = 1 2 = cos  3   = 2n   3 ; n  Z  cÖ`Ë e ̈ewai g‡a ̈ gvb  = 0,  3  Ackb Abyhvqx DËi =  3 A_ev, Option Test: Ackb 5 3 , 0 ≤  <  2 e ̈ewai AšÍfz©3 bq| Ackb  =  3 L.H.S = tan 3 + cot  3 = 3 + 1 3 = 4 3 R.H.S = 2cosec 3 = 2  2 3 = 4 3  Ackb mwVK DËi| 4. cosec2     tan–1 1 2 – 3sec2 (cot ) –1 3 = ? [DU 21-22] 15 2 9 1 25 DËi: 1
2  Higher Math 2nd Paper Chapter-7 e ̈vL ̈v: cosec2     tan–11 2 – 3sec2 (cot ) –1 3 2 1 2 3 1 5 = {cosec(cosec )} –1 5 2 – 3       sec    sec–1 2 3 2 = 5 – 3  4 3 = 5 – 4 = 1 5. tan–1 2 3 + cos–1 2 13 = ? [DU 18-19] tan–1     5 9 tan–1     3 7  2  4 DËi:  2 e ̈vL ̈v: tan–1 2 3 + cos–1 2 13 = tan–1 2 3 + tan–1 3 2 = tan–1 2 3 + cot–1 2 3 =  2 3 2 13 6. cot cot3 = 1 mgxKi‡Yi mvaviY mgvavbÑ [DU 18-19] (2n + 1) 4 (2n + 1) 8 n 4 (2n + 1) 2 DËi: (2n + 1) 8 e ̈vL ̈v: cot cot3 = 1  cot cot3 – 1 = 0  cot cot3 – 1 cot + cot3 = 0  cot4 = 0 = cot 2  4 = n +  2   = n 4 +  8   = (2n + 1) 8 ; n  Z 7. sin–1 x + sin–1 y =  2 n‡j, †KvbwU mwVK? [DU 17-18; KU 17-18] x 2 + y2 = 1 x 2 – y 2 = 1 x + y = 1 x – y = 1 DËi: x 2 + y2 = 1 e ̈vL ̈v: sin–1 x + sin–1 y =  2  sin–1 x =  2 – sin–1 y  x = sin     2 – sin–1 y  x = cos(sin ) –1 y  x = 1 – y 2  x 2 = 1 – y 2  x 2 + y2 = 1 8. 0  x  90 n‡j, sin3x = cosx mgxKi‡Yi mgvavb n‡eÑ [DU 17-18] 0, 45 0, 22.5 45, 45 22.5, 45 DËi: 22.5, 45 e ̈vL ̈v: sin3x = cosx  cos     2 – 3x = cosx  x = 2n       2 – 3x (+) wb‡q, x = 2n +  2 – 3x  4x = 2n +  2  x = n 2 +  8  0  x  90 e ̈ewai g‡a ̈, x = 22.5, 45 (–) wb‡q, x = 2n –   + 3x  – 2x = 2n –    2x =  2 – 2n  x =  4 – n 9. hw` tan–1     x + 1 3 + tan–1    x –  1 3 = tan–1 2 nq, Z‡e x Gi gvb n‡eÑ [DU 16-17] – 5 6 – 1 3 1 3 2 3 DËi: 2 3 e ̈vL ̈v: tan–1     x + 1 3 + tan–1    x –  1 3 = tan–1 2  tan–1 x + 1 3 + x – 1 3 1 –     x 2 – 1 9 = tan–1 2
wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY  Varsity Practice Sheet Solution 3  tan–1 2x 10 9 – x 2 = tan–1 2  2x = 2    10 9 – x 2  x = 10 9 – x 2  9x2 + 9x – 10 = 0  x = – 9  9 2 – 4  9(– 10) 2  9  x = – 9  441 18  x = – 9  21 18  x = 2 3 [+ ve wPý wb‡q] A_ev, Option Test: x = 2 3 ewm‡q, tan–1     x + 1 3 + tan–1    x –  1 3 = tan–1 2x 10 9 – x 2 = tan–1 4 3 10 9 – 4 9 = tan–1     4 3  9 6 = tan–1 2 10. sec2  + tan2  = 5 3 ; 0 <  <  Gi mgvavb †KvbwU? [DU 16-17; RU 21-22]  3 , 5 6 –  6 , 5 6  6 , – 5 6  6 , 5 6 DËi:  6 , 5 6 e ̈vL ̈v: sec2  + tan2  = 5 3  1 + tan2  + tan2  = 5 3  2tan2  = 2 3  tan2  = 1 3  tan =  1 3 = tan      6   = n   6 ; n  Z   =  6 , 5 6 [⸪ 0    ] 11. cot2  – ( 3 + 1)cot + 3 = 0, 0 <  <  2 ,  = KZ? [DU 15-16; JU 18-19]  6 ,  3  4 ,  3  3 ,  5  6 ,  4 DËi:  6 ,  4 e ̈vL ̈v: cot2  – ( 3 + 1)cot + 3 = 0  cot2  – 3cot – cot + 3 = 0  cot(cot – 3) – 1(cot – 3) = 0  (cot – 1)(cot – 3) = 0 nq, cot – 1 = 0  cot = 1  tan = 1 = tan  4   = n +  4 ; n  Z A_ev, cot – 3 = 0  cot = 3  tan = 1 3 = tan  6   = n +  6 ; n  Z   =  6 ,  4     ⸪ 0     2 A_ev, Option Test K‡iv| 12. sec2 (cot–1 3) + cosec2 (tan–1 2) = ? [DU 15-16] 85 36 36 85 10 9 9 10 DËi: 85 36 e ̈vL ̈v: sec2 (cot–1 3) + cosec2 (tan–1 2) = 1 + tan2     tan–11 3 + 1 + cot2     cot–11 2 = 2 + 1 9 + 1 4 = 85 36 13. cosec + cot = 3, (0 <  < 2) n‡j  Gi gvbÑ [DU 14-15; IU 16-17]  6  4  3 2 3 DËi:  3 e ̈vL ̈v: cosec + cot= 3  1 sin + cos sin = 3  cos – 3sin + 1 = 0
4  Higher Math 2nd Paper Chapter-7  cos – 3sin = – 1  1 2 cos – 3 2 sin = – 1 2  cos     3 +  = cos 2 3   3 +  = 2 3   =  3 A_ev, Option Test K‡iv| 14. sin–1 4 5 + cos–1 2 5 mgvbÑ [DU 14-15] tan–1 2 11 sin–1 11 2 tan–1 11 2 cos–1 11 2 DËi: tan–1 11 2 e ̈vL ̈v: sin–1 4 5 + cos–1 2 5 = tan–14 3 + tan–11 2 = tan–1 4 3 + 1 2 1 – 4 3  1 2 = tan–1 11 6 6 – 4 6 = tan–111 2 5 3 5 1 2 4 15. arc tan      sin       arc cos 2 3 mgvbÑ [DU 13-14]  2  3  4  6 DËi:  6 e ̈vL ̈v: arc Øviv inverse †evSvq arc tan      sin       arc cos 2 3 = tan–1       sin      cos–1 2 3 2 1 3 = tan–1       sin    sin–1 1 3 = tan–1 1 3 =  6 16. cot + 3 = 2cosec mgxKi‡Yi mgvavbÑ [DU 13-14]  = 2n –  3  = 2n +  3  = 2n +  6  = 2n –  6 DËi:  = 2n +  3 e ̈vL ̈v: cot + 3 = 2cosec  cos sin + 3 = 2 sin  cos + 3sin = 2  1 2 cos+ 3 2 sin = 1  cos     –  3 = 1   –  3 = 2n   = 2n +  3 ; n  Z 17. cotx – tanx = 2 mgxKiYwUi mvaviY mgvavbÑ [DU 05-06] n 4 n 2 (4n + 1) 8 (4n + 1) 2 DËi: (4n + 1) 8 e ̈vL ̈v: cotx – tanx = 2  cosx sinx – sinx cosx = 2  cos2 x – sin2 x sinx cosx = 2  cos2x = sin2x  sin2x cos2x = 1  tan2x = tan  4  2x = n +    x = n 2 +  8  x = (4n + 1) 8 ; n  Z weMZ mv‡j GST-G Avmv cÖkœvejx 1. sin–1 p + sin–1 q =  2 n‡j, p 1 – q 2 + q 1 – p 2 Gi gvb KZ? [GST 23-24] – 1 0 1 2 DËi: 1