Tiêu đề Trigonometry Varsity Practice Sheet Solution.pdf ✅

mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Varsity Practice Sheet Solution 1 07 mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ Trigonometric Ratios of Associated Angles weMZ mv‡j DU-G Avmv cÖkœvejx 1. tan75 Gi gvb KZ? [DU 23-24; JU 14-15] 1 – 3 1 + 3 3 + 1 3 – 1 3 – 1 3 + 1 1 + 3 1 – 3 DËi: 3 + 1 3 – 1 e ̈vL ̈v: tan75 = tan(30 + 45) = tan30 + tan45 1 – tan30tan45 = 1 3 + 1 1 – 1 3  1 = 1 + 3 3 – 1 = 3 + 1 3 – 1 2. cosec10 – 4sin70 Gi gvb KZ? [DU 22-23] – 1 1 2 – 2 2 DËi: 2 e ̈vL ̈v: cosec10 – 4sin70 = 1 sin10 – 4sin70 = 1 – 4sin70sin10 sin10 = 1 – 2(cos60 – cos80) sin10 = 1 – 2  1 2 + 2cos80 sin10 = 2sin10 sin10 = 2 3. tan + sec = x n‡j, cosec Gi gvb KZ? [DU 22-23] x 2 + 1 x 2 – 1 x 2 – 1 x 2 + 1 1 – x 2 1 + x2 1 + x2 1 – x 2 DËi: x 2 + 1 x 2 – 1 e ̈vL ̈v: tan + sec = x  sin cos + 1 cos = x  sin + 1 cos = x  2sin 2 cos  2 + sin2 2 + cos2 2     cos2 2 – sin2 2 = x      cos  2 + sin 2 2     cos  2 – sin 2     cos  2 + sin 2 = x      cos  2 + sin 2 2     cos  2 – sin 2 2 = x 2  1 + sin 1 – sin = x 2  1 + sin + 1 – sin 1 + sin – 1 + sin = x 2 + 1 x 2 – 1 [†hvRb-we‡qvRb K‡i]  1 sin = cosec = x 2 + 1 x 2 – 1 4. cos2 (60 + A) + cos2 (60 – A) Gi gvbÑ [DU 18-19] 1 – 1 2 cos2A 1 + sin2A 1 + 3 cos2A 1 + 1 2 cos2A DËi: 1 – 1 2 cos2A

mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Varsity Practice Sheet Solution 3 11. sin65 + cos65 Gi gvbÑ [DU 14-15] 2cos20 2cos20 2sin20 2sin20 DËi: 2cos20 e ̈vL ̈v: sin65 + cos65 = sin(90 – 25) + cos65 = cos25 + cos65 = 2cos    65 + 25 2 cos    65 – 25 2 = 2cos45cos20 = 2  1 2 cos20 = 2cos20 12. ABC wÎfz‡Ri cosA + cosC = sinB n‡j, C Gi gvbÑ [DU 14-15]  4  3  2  6 DËi:  2 e ̈vL ̈v: cosA + cosC = sinB  2cos A + C 2 cos A – C 2 = 2sinB 2 cos B 2  cos  – B 2 cos C – A 2 = sinB 2 cos B 2  sinB 2 cos C – A 2 = sinB 2 cos B 2    ⸪ cos  A – C 2 = cos C – A 2  C = A + B  A + B + C =   C + C =   C =  2 13. GKK e ̈vmv‡a©i e„‡Ë AšÍwj©wLZ GKwU mgevû wÎfz‡Ri evûi •`N© ̈Ñ [DU 13-14] 3 2 units 3 2 units 3 units 1 unit DËi: 3 units e ̈vL ̈v: Avgiv sine m~Î n‡Z Rvwb, a sinA = 2R  a = 2  1  sinA [⸪ R = 1; GKK e ̈vmva© e„Ë] 60 R  a = 2sin60 [⸪ mgevû wÎfzR ZvB A = 60]  a = 2  3 2 = 3  a = 3 14. sin75 + sin15 sin75 – sin15 Gi gvbÑ [DU 11-12, 04-05; RU 15-16; JnU 10-11] 5 3 – 3 – 5 DËi: 3 e ̈vL ̈v: sin75 + sin15 sin75 – sin15 = sin(90 – 15) + sin15 sin(90 – 15) – sin15 = cos15 + sin15 cos15 – sin15 = 1 + sin15 cos15 1 – sin15 cos15 = 1 + tan15 1 – tan15 = tan45 + tan15 1 – tan45 . tan15 = tan(45 + 15) = tan60 = 3 15. 2tan 1 + tan2  = ? [DU 01-02, 98-99; IU 16-17; CU 12-13; RU 12-13;] tan2 2sincos 2cos2  2 cos2 DËi: 2sincos e ̈vL ̈v: 2tan 1 + tan2  = sin2 = 2sincos weMZ mv‡j GST-G Avmv cÖkœvejx 1.  = cos–1     4 5 n‡j, (1 – tan2 ) (1 + tan2 ) = ? [GST 23-24; KU 17-18; JU 14-15; RU 09-10] 16 25 9 25 8 25 7 25 DËi: 7 25 e ̈vL ̈v: 1 – tan2  1 + tan2  = cos2 = 2cos2  – 1 = 2  16 25 – 1 = 32 – 25 25 = 7 25
4  Higher Math 1st Paper Chapter-7 2. sin4 x + cos4 x Gi ch©vq KZ? [GST 23-24]  2  3  2 3 DËi:  2 e ̈vL ̈v: sin4 x + cos4 x Gi ch©vq = 2 4 =  2 3. sin      3 – sin–1    –  1 2 = ? [GST 23-24] 1 2 1 3 – 1 1 DËi: 1 e ̈vL ̈v: sin      3 – sin–1    –  1 2 = sin      3 +  6 = sin  2 = 1 4. 2cos2 x + 3cosx = 2, 0  x  2 Gi mgvavb †mU- [GST 22-23]        3  5         3          3  2         2  5  DËi:        3  5  e ̈vL ̈v: 2cos2 x + 3cosx = 2  2cos2 x + 3cosx – 2 = 0  2cos2 x + 4cosx – cosx – 2 = 0  2cosx(cosx + 2) – 1(cosx + 2) = 0  (cosx + 2) (2cosx – 1) = 0  cosx = 1 2 [⸪ cosx  – 2]  x = 2n   3 0  x  2 e‡j, x =        3  5 3 5. in     –  6 + sin     + 5 6 Gi gvb KZ? [GST 22-23] – 1 0 – cos 3sin DËi: 0 e ̈vL ̈v: sin     –  6 + sin     + 5 6 = sin     –  6 + sin     +  –  6 = sin     –  6 + sin       –     –  +  6 = sin     –  6 – sin     –  6 = 0 6. [– , 2 e ̈ewa‡Z cos + 1 = 0 Gi mgvavb †mU †KvbwU? [GST 21-22] {– , } {, 2}        3        –  3  DËi: {– , } e ̈vL ̈v: cos + 1 = 0  cos = – 1   = {– , } [⸪ e ̈ewa [– , 2]] Note: –  I  GKB †KvY †evSvq hv x A‡ÿi FYvZ¥K †iLv eivei wb‡`©k K‡i| †h‡nZz e ̈ewa [– , 2]; ZvB –  †K Avevi D‡jøL Ki‡Z n‡e| 7. hw` cosx + cosy = p Ges sinx + siny = q nq Z‡e, tan x + y 2 = ? [GST 20-21] p q q p p 2 q 2 q 2 p 2 DËi: q p e ̈vL ̈v: cosx + cosy = p  2cos x + y 2 cos x – y 2 = p ......... (i) sinx + siny = q  2sin x + y 2 cos x – y 2 = q ......... (ii) (ii)  (i) K‡i, tan x + y 2 = q p 8. hw` cos + sec = 2 nq, Z‡e cos3  + sec3  Gi gvb KZ? [GST 20-21] 1 2 3 8 DËi: 2 e ̈vL ̈v: cos + sec  cos2  + 1 = 2cos  cos2  – 2cos + 1 = 0  (cos – 1)2 = 0  cos = 1  sec = 1  cos3  + sec3    weMZ mv‡j Agri-G Avmv cÖkœvejx 1. A + B =  4 n‡j, (1 + tanA)(1 + tanB) Gi gvb KZ? [Agri. Guccho 20-21; RU 13-14; JU 22-23] 1 2 3 3 3 DËi: 2