Tiêu đề Inverse Trigonometric Engg Practice Sheet Solution.pdf ✅

wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY  Engineering Question Bank Solution 1 07 wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY Inverse Trigonometric Functions and Trigonometric Equations WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1| mgvavb Ki: cos3 x – 1 2 sin2x = sin3 x + 1 [BUET 22-23] mgvavb: cos3 x – sin3 x = 1 + 1 2 sin2x  (cosx – sinx) (cos2 x + sin2 x + sinxcosx) = 1 + sinxcosx  (cosx – sinx) (1 + sinxcosx) – (1 + sinxcosx) = 0  (1 + sinxcosx)(cosx – sinx – 1) = 0 nq, 1 + sinxcosx = 0  sinxcosx = –1  2sinxcosx = – 2  sin2x = – 2 wKš‘, sin2x  – 2 A_ev, cosx – sinx = 1  1 2 cosx – 1 2 sinx = 1 2  cos     x +  4 = cos  4  x +  4 = 2n   4  x = 2n, 2n –  2 ; n  Z (Ans.) 2| log4 (sinx – cos2x)3 = 2       tan–1–5 2 + tan–13 7 n‡j x Gi gvb wbY©q Ki| hLb [– 2  x  2] [BUET 21-22] mgvavb: †`Iqv Av‡Q, log4 (sinx – cos2x)3 = 2       tan–1–5 2 + tan–13 7  3 log4 (sinx – cos2x) = 2 tan–1 (– 1)  3 log4 (sinx – cos2x) = 2 × 3 4  log4 (sinx – cos2x) = 1 2  sinx – cos2x = 4 1 2  sinx – cos2x = 2  sinx + 2sin2 x – 1 = 2 [∵ 1 – cos2x = 2sin2 x]  2 sin2 x + sinx – 3 = 0  2 sin2 x + 3 sinx – 2 sinx – 3 = 0  sinx (2 sinx + 3) – 1(2 sinx + 3) = 0  (sinx – 1) (2 sinx + 3) = 0 nq, sinx – 1 = 0 A_ev, 2 sinx + 3 = 0  sinx = 1  sinx = – 3 2 [wKš‘ sinx  – 3 2 ]  x = (4n + 1)  2  [– 2  x  2] mxgvi g‡a ̈ MÖnY‡hvM ̈ gvb ̧‡jv n‡jv: x =  2 , – 3 2 (Ans.) 3| sin(x) + sin     x 2 = 0 n‡j, x = ? [0  x  2] [BUET 20-21] mgvavb: sin(x) + sin x 2 = 0  2sinx 2  cos x 2 + sinx 2 = 0  sinx 2     2cos x 2 + 1 = 0  sinx 2 = 0  x 2 = n  x = 2n n = 0 n‡j, x = 0 n = 1 n‡j, x = 2 Avevi, cos x 2 = – 1 2  cos x 2 = cos 2 3  x 2 = 2n  2 3  x = 4n  4 3 n = 0 n‡j, x =  4 3    ∵  – 4 3 MÖnY‡hvM ̈ bq  wb‡Y©q gvb, x = 0, 4 3 , 2 (Ans.) 4| atan + bsec = c mgxKi‡Yi g~jØq ,  n‡j cÖgvY Ki †h, tan( + ) = 2ac a 2 – c 2| [BUET 19-20; MIST 19-20] mgvavb: atan + bsec = c  atan – c = – bsec  a 2 tan2  – 2catan + c2 = b2 sec2  [eM© K‡i]  a 2 tan2  – 2catan + c2 = b2 (1 + tan2 )  a 2 tan2  – 2catan + c2 – b 2 – b 2 tan2  = 0  (a2 – b 2 )tan2  – 2catan + (c 2 – b 2 ) = 0
2  Higher Math 2nd Paper Chapter-7 GLb, tan + tan = 2ca a 2 – b 2 ; tantan = c 2 – b 2 a 2 – b 2 [ax2 + bx + c = 0 mgxKi‡Yi g~jØq ,  n‡j,  +  = – b a Ges  = c a ] L.H.S = tan( + ) = tan + tan 1 – tantan = 2ca a 2 – b 2 1 – c 2 – b 2 a 2 – b 2 = 2ca a 2 – c 2 = R.H.S (Proved) 5| mgvavb Ki: sin–1 2x + sin–1 x =  3 [BUET 18-19; MIST 18-19] mgvavb: sin–1 2x + sin–1 x =  3  sin–1 2x =  3 – sin–1 x  2x = sin      3 – sin–1 x  2x = 3 2 1 – x 2 – 1 2 x [sin(A – B) = sinAcosB – cosA sinB]  5x 2 = 3 2 1 – x 2  25x2 = 3 – 3x2  28x2 = 3  x =  3 28 ïw× cixÿv K‡i cvB, x = 3 28 (Ans.) 6| mgvavb Ki: 2(sinxcosx + 3) = 3cosx + 4sinx, 0 < x <  [BUET 17-18; MIST 16-17] mgvavb: 2sinxcosx + 2 3 – 3cosx – 4sinx = 0  2sinx(cosx – 2) – 3(cosx – 2) = 0  (2sinx – 3) (cosx – 2) = 0  2sinx – 3 = 0 A_ev, cosx – 2 = 0 wKš‘, cosx  2  sinx = 3 2 = sin  3  x = n + (–1)n  3 [†hLv‡b, n  Z] n = 0 n‡j, x =  3 ; n = 1 n‡j, x =  –  3 = 2 3 x = 2 n‡j, x = 2 +  3 = 7 3  wb‡Y©q mgvavb x =  3  2 3 (Ans.) 7| mgvavb Ki: cos–1 x – sin–1 x = sin–1 (1 – x) [BUET 17-18] mgvavb: 1 x 1 – x 2 †`Iqv Av‡Q, cos–1 x – sin–1 x = sin–1 (1 – x)  sin–1 ( 1 – x 2 ) – sin–1 (x) = sin–1 (1 – x)  sin–1 (1 – x 2 – x 1 – 1 + x 2 ) = sin–1 (1 – x)  sin–1 (1 – x 2 – x 2 ) = sin–1 (1 – x)  sin–1 (1 – 2x2 ) = sin–1 (1 – x)  1 – 2x2 = 1 – x  2x2 = x  2x2 – x = 0  x(2x – 1) = 0  x = 0 A_ev, x = 1 2 wb‡Y©q mgvavb, x = 0, 1 2 (Ans.) 8| mgvavb Ki: cotx + cot2x + cot3x = cotx cot2x cot3x [BUET 14-15] mgvavb: cotx + cot2x + cot3x = cotx cot2x cot3x  cot2x + cotx = cot3x (cotx cot2x – 1)  cot2x + cotx cotx cot2x – 1 = cot3x  1 cot(x + 2x) = cot3x  tan3x = 1 tan3x  tan 2 3x = 1  tan3x =  1 = tan       4  3x = n   4  x = n 3   12 ; [n  Z] (Ans.) 9| tan–1 x + 2cot–1 x = 2 3  [BUET 10-11] mgvavb: tan–1 x + cot–1 x + cot–1 x = 2 3    2 + cot–1 x = 2 3   cot–1 x =  6 = cot–1 3 x = 3 (Ans.) 10| mgvavb Ki: 4cosx cos2x cos3x = 1; 0 < x <  [BUET 08-09] mgvavb: 4cosx cos2x cos3x = 1  2cos2x (cos4x + cos2x) = 1  2cos2x cos4x + 2cos2 2x – 1 = 0  2cos2x cos4x + cos4x = 0  cos4x (2cos2x + 1) = 0
wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY  Engineering Question Bank Solution 3 nq, cos4x = 0 4x = (2n + 1)  2  x = (2n + 1) 8 A_ev, 2cos2x + 1 = 0  cos2x = –1 2 = cos 2 3  2x = 2n  2 3 x = n   3  x =  8 ,  3 , 3 8 , 2 3 , 5 8 , 7 8 (Ans.) 11| mgvavb Ki: sin + sin2 + sin3 = 1 + cos + cos2; 0 <  <  [BUET 08-09] mgvavb: sin + sin2 + sin3 = 1 + cos + cos2  2sin2 cos + sin2 = 2 cos2  + cos  sin2 (2cos + 1) = cos (1 + 2 cos)  (1 + 2cos).(sin2 – cos) = 0 nq, 1 + 2cos = 0  cos = – 1 2  cos = cos 2 3   = 2n  2 3 A_ev, sin2 – cos = 0 2 sin cos – cos = 0  cos (2 sin – 1) = 0  cos = 0 Avevi, sin = 1 2   = (2n + 1) 2  sin = sin  6   = n + (– 1)n  6  = (2n + 1)  2 , n + (– 1)n  2 , 2n  2 3 †hLv‡b, n  Z  0 <  <  e ̈ewa‡Z  Gi MÖnY‡hvM ̈ gvb:  2 ,  6 , 2 3 , 5 6 (Ans.) 12| mgvavb Ki: 3sinx – cosx = 2, 0 < x < 2 [BUET 07-08] mgvavb: 3 sinx – cosx = 2 3 2 sinx – 1 2 cosx = 1 [Dfqcÿ‡K 2 Øviv fvM K‡i]  sinx cos  6 – cosx sin  6 = 1  sin     x –  6 = 1  x –  6 = (4n + 1)  2  x = (4n + 1)  2 +  6 ; †hLv‡b, n  Z  wb‡Y©q mgvavb : x = 2 3 (Ans.) 13| mgvavb K‡iv: tan–1 1 – x 1 + x = 1 2 tan–1 x [BUET 06-07] mgvavb: tan–1 1 – x 1 + x = 1 2 tan–1 x  tan–1 1 – tan–1 x = 1 2 tan–1 x  tan–1 1 = 3 2 tan–1 x =  4  tan–1 x = 2 12  x = tan  6 = 1 3 (Ans.) 14| mgvavb K‡iv: 3 sin 2x – 1 cos 2x = 4 [BUET 06-07; MIST 17-18] mgvavb: 3 sin2x – 1 cos2x = 4  3 cos2x – sin2x sin2x cos2x = 4  3 2 cos2x – 1 2 sin2x = 2sin2x cos2x sin  3 cos2x – cos  3 sin2x = 2sin2x cos2x  sin      3 – 2x = sin4x  – sin     2x –  3 = sin4x  sin4x + sin     2x –  3 = 0  2sin     3x –  6 cos     x +  6 = 0 nq, sin     3x –  6 = 0  3x –  6 = n x = (6n + 1)  18 A_ev, cos    x +  6 = 0  x +  6 = (2n + 1)  2  x = (3n + 1)  3 wb‡Y©q mgvavb: x = (6n + 1)  18 , (3n + 1)  3 ; n  Z (Ans.) 15| mgvavb wbY©q Ki: 1 – 2sin = cos [BUET 05-06] mgvavb: cos + 2 sin = 1  1 5 cos + 2 5 sin = 1 5  cos cos + sin sin = cos  cos( – ) = cos   –  = 2n     = 2n + 2, 2n †hLv‡b, n  Z Ges  = cos–1 1 5 awi, 1 5 = cos  sin = 2 5 5 1 2 
4  Higher Math 2nd Paper Chapter-7 16| mgvavb Ki: 1 + sin2 + sin2 = cos(2 + 2) [0  ,   90] [BUET 05-06] mgvavb: 1 + sin2 + sin2 = cos(2 + 2)  2sin( + ) cos( – ) + 1 – cos2( + ) = 0  2sin( + ) cos( – ) + 2sin2 ( + ) = 0  2sin( + ) {cos( – ) + sin( + )} = 0  2sin( + ) = 0   +  = n  0  ,   90   +  = 0, 180   =  = 0 A_ev,  =  = 90 cos( – ) + sin( + )  0 [0  ,   90 Gi g‡a ̈  I  Gi †Kv‡bv gv‡bi R‡b ̈B cos( – ) = 0 n‡Z cv‡i bv] 17| mgvavb Ki: cos + sin = 2 [BUET 02-03] mgvavb: cos + sin = 2  1 2 cos + 1 2 sin = 2 2  sin  4 cos + cos  4 sin = 1  sin      4 +  = 1   4 +  = (4n + 1)  2   = (4n + 1)  2 –  4 = 2n +  4 ; n  Z (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 18| cÖgvY Ki †h, costan–1 cotsin–1 x = x (wPÎ Avek ̈K) [KUET 19-20] mgvavb: L.H.S = cos tan–1 cot sin–1 x = cos tan–1 cot cot–1 1 – x 2 x x  1 – x 2 1 = cos tan–1 1 – x 2 x = cos cos–1 x 1 = x = R.H.S (Proved) x  1 – x 2 1 19| cÖgvY Ki: cos–1 x = 2sin–1 1 – x 2 = 2cos–1 1 + x 2 [KUET 03-04] mgvavb: awi, cos–1 x =   cos = x GLb, sin 2 = 1– cos 2 = 1 – x 2   2 = sin–1 1 – x 2   = 2sin–1 1 – x 2  cos–1 x = 2sin–1 1 – x 2 Avevi, cos  2 = 1+ cos 2 = 1 + x 2   2 = cos–1 1 + x 2   = 2cos–1 1 + x 2  cos–1 x = 2cos–1 1 + x 2  cos –1 x = 2sin–1 1 – x 2 = 2cos–1 1 + x 2 20| mgvavb Ki: tan2  = 3cosec2  – 1 for 0    2 [KUET 03-04] mgvavb: tan2  = 3cosec2  – 1  1 + tan2  = 3cosec2   sec 2  = 3cosec2   sin2  cos 2  = 3  tan2  = 3  tan =  3 GLb, tan = 3 n‡j tan = tan  3   = n +  3 Avevi, tan = – 3 n‡j tan = – tan  3   = n –  3   =  3 , 4 3 , 2 3 , 5 3 weMZ mv‡j RUET-G Avmv cÖkœvejx 21| hw` tan–1 (x + ) – tan–1 x =  4 nq Z‡e x I -Gi g‡a ̈ m¤úK© wbY©q Ki Ges G‡`i m¤¢ve ̈ gvb †ei Ki| (x > 0,  > 0) [RUET 19-20; MIST 15-16] mgvavb: tan–1 (x + ) – tan–1 x =  4  tan–1 x +  – x 1 + x(x + ) = tan–1 (1)   1 + x2 + x = 1   = x 2 + x + 1  x 2 + x + 1 –  = 0  x = –    2 – 4  1  (1 – ) 2  1 x I  Gi m¤úK©, x = –    2 + 4 – 4 2 (Ans.)