Tiêu đề 6. Trigonometric Ratios Engineering Practice Sheet Solution [HSC 26].pdf ✅

2  Higher Math 1st Paper Chapter-6 2. hw` 3sec4  + 8 = 10sec2  nq, Z‡e tan Gi gvb n‡eÑ [BUET 12-13]  1 3  1  1 2  1  1 3 ,  1 DËi:  1 3 ,  1 e ̈vL ̈v: 3sec4  – 10sec2  + 8 = 0  (sec2  – 2) (3sec2  – 4) = 0 sec2  – 2 = 0  1 + tan2  – 2 = 0  tan =  1 3sec2  – 4 = 0  3 + 3tan2  – 4 = 0  3tan2  = 1  tan =  1 3 3. hw` cot2  + cosec – 5 = 0 nq, ZLb  abvZ¥K, Zvn‡j 0 <  <  2 Gi Rb ̈  Gi gvb n‡e: [BUET 09-10] 0 30 45 60 DËi: 30 e ̈vL ̈v: cot2  + cosec – 5 = 0  cosec2  + cosec – 1 – 5 = 0  cosec2  + cosec – 6 = 0  (cosec + 3)(cosec – 2) = 0  cosec = – 3, cosec = 2; 0 <  <  2   = 30 weMZ mv‡j KUET-G Avmv cÖkœvewj 4. hw` sin + cosec = 2 nq Z‡e sinn  + cosecn  Gi gvb n‡jvÑ [KUET 12-13] 1 – 1 2 – 2 3 DËi: 2 e ̈vL ̈v: sin + cosec = 2  sin + 1 sin = 2  sin2  + 1 = 2sin  sin2  – 2 sin + 1 = 0  (sin – 1)2 = 0  sin – 1 = 0  sin = 1  cosec = 1 sin = 1 1 = 1  sinn  + cosecn  = 1n + 1n = 1 + 1 = 2 5. sin †K cot Gi gva ̈‡g cÖKvk Ki| [KUET 06-07]  1 1 + cot2  1 2 + cot2  1 1 + cot2   1 1 – cot2  DËi:  1 1 + cot2  e ̈vL ̈v: sin = 1 cosec =  1 1 + cot2  6. 5 cm e ̈vmva© wewkó e„‡Ëi GKwU Pvc †K‡›`a 40 †KvY Drcbœ Ki‡j H Pv‡ci •`N© ̈ KZ cm? [KUET 19-20] 3.491 3.520 3.641 4.00 DËi: 3.491 e ̈vL ̈v: GLv‡b,  = 40 = 40   180 = 2 9  s = r = 5  2 9 = 3.491 weMZ mv‡j CUET-G Avmv cÖkœvewj 7. cos–2     144 169 =  n‡j tan Gi gvb n‡eÑ [CUET 09-10] 25/169 169/25 25/144 None of them DËi: None of them e ̈vL ̈v: cos–2     144 169 =   cos2  = 144 169  sec2  = 169 144  1 +tan2  = 169 144  tan2  = 25 144  tan =  5 12 8. cos2  = (a + b) 2 4ab mgxKi‡Y a = b n‡j  Gi gvb n‡eÑ [CUET 09-10] 0 90 45 None of them DËi: 0