Tiêu đề Trigonometric Ratios Practice Sheet Solution for Academic Batch.pdf ✅

mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Final Revision Batch '24 1 07 mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ Trigonometric Ratios of Associated Angles Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 2 1 1 2 1 1 1 2 1 2022 2 1 2 1 1 1 1 2 1 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 4 4 5 4 5 5 4 4 4 2022 4 4 2 4 4 4 4 4 4 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| C A  120 40 B [XvKv †evW©- Õ23] (K) cot = 2 n‡j, cos2 Gi gvb wbY©q Ki| (L) DÏxc‡Ki Av‡jv‡K †`LvI †h, 3 4 cosecA – 1 4 secA = 1 (M) DÏxc‡Ki Av‡jv‡K †`LvI †h, 3 – cos2 ( + A) – cos2A – cos2 ( – A) = 3 2 mgvavb: (K) †`Iqv Av‡Q, cot = 2  tan = 1 2 GLb, cos2 = 1 – tan2  1 + tan2  = 1 – 1 2 1 + 1 2 = 1 2 3 2 = 1 3 (Ans.) (L) ABC wÎfz‡R, B = 40, C = 120 C A  120 40 B  A + B + C = 180 = 180 – 40 – 120 = 20 L.H.S = 3 4 cosecA – 1 4 secA = 1 4 ( 3 cosec20 – sec20) = 1 4    3  sin20 – 1 cos20 = 1 4    3 cos20 – sin20 sin20.cos20 = 1 2     3  2 cos20 – 1 2 sin20 sin20.cos20 = sin60.cos20 – cos60.sin20 2sin20.cos20 = sin(60 – 20) sin(2.20) = sin40 sin40 = 1 = R.H.S (Showed) (M) wPÎ n‡Z, A = 20 C = 120 GLb,  = 180 – C = 120 C A  120 40 B L.H.S = 3 – cos2 ( + A) – cos2A – cos2 ( – A) = 3 – cos2 80 – cos2 20 – cos2 40 = 3 – 1 2 (2cos2 80 + 2cos2 20 + 2cos2 40) = 3 – 1 2 (1 + cos160 + 1 + cos40 + 1 + cos80) = 3 – 1 2 (3 + cos40 + cos80 + cos160) = 3 – 1 2 – 1 2 {cos40 + (cos80 + cos160)}
2  Higher Math 1st Paper Chapter-7 = 3 2 – 1 2 (cos40 + 2.cos120 . cos40) = 3 2 – 1 2 (cos40 – cos40) = 3 2 = R.H.S (Showed) 2| `„k ̈Kí-1: A B C `„k ̈Kí-2: cosx + cosy = a, sinx + siny = b [XvKv †evW©- Õ23] (K) tan3 †K tan Gi gva ̈‡g cÖKvk Ki| (L) `„k ̈Kí-2 Gi Av‡jv‡K cos(x + y) Gi gvb a I b Gi gva ̈‡g cÖKvk Ki | (M) `„k ̈Kí-1 Gi Av‡jv‡K cÖgvY Ki †h, BC cosC – BC cosB = (AC – AB) (1 + cosA) mgvavb: (K) tan3 = tan( + 2) = tan + tan2 1 – tan.tan2 = tan+ 2tan 1 – tan2  1 – tan 2tan 1 – tan2  = tan – tan3  + 2tan 1 – tan2  1– tan2  – 2tan2  1 – tan2  = 3tan – tan3  1 – 3 tan2  (Ans.) (L) †`Iqv Av‡Q, cosx + cosy = a  2cos x + y 2 .cos x – y 2 = a ....(i) Ges sinx + siny = b  2sinx + y 2 .cos x – y 2 = b .....(ii) (i)  (ii) K‡i cvB,  cos x + y 2 sinx + y 2 = a b  2cos2 x + y 2 2sin2 x + y 2 = a 2 b 2 [eM© Kivi ci 2 Øviv ̧Y]  1 + cos(x + y) 1 – cos(x + y) = a 2 b 2  1 + cos(x + y) – 1 + cos(x + y) 1 + cos(x + y) + 1 – cos(x + y) = a 2 – b 2 a 2 + b2 [we‡qvRb-†hvRb K‡i]  2cos(x + y) 2 = a 2 – b 2 a 2 + b2  cos(x + y) = a 2 – b 2 a 2 + b2 (Ans.) (M) A B C c b a wPÎ n‡Z, AB = c, BC = a, AC = b Ges A + B + C =  R.H.S = (AC – AB) (1 + cosA) = (b – c).2cos2A 2 = 2a. b – c a .cos2A 2 = 2acos2A 2 2R sinB – 2R sinC 2R sinA = 2a cos2A 2 sinB – sinC sinA = 2a cos 2A 2 2cos B + C 2 .sinB – C 2 sinA = 2acos2A 2 2cos  – A 2 .sinB – C 2 2sinA 2 .cos A 2 [⸪ A + B + C = ] = 2acos A 2 sinA 2 .sinB – C 2 sinA 2 = 2acos A 2 sinB – C 2 = 2a.sinB – C 2 cos  – (B + C) 2 = 2a sinB + C 2 . sinB – C 2 = a(cosC – cosB) = BC cosC – BC cosB = L.H.S (Proved) 3| B A x 12 km 45 N 9 km O Q l R P T = l + m + n n m `„k ̈Kí-1: `„k ̈Kí-2: N ON || BN [ivRkvnx †evW©- Õ23] (K) tan = y x n‡j, †`LvI †h, x cos2 + y sin2 = x (L) `„k ̈Kí-1 e ̈envi K‡i AB `~iZ¡ wbY©q Ki| (M) 1 T – m + 1 T – l = 3 T n‡j, `„k ̈Kí-2 Gi wÎfz‡Ri R †KvY wbY©q Ki|
mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Higher Math Academic Batch 3 mgvavb: (K) †`Iqv Av‡Q, tan = y x L.H.S = x cos2 + ysin2 = x. 1 – tan2  1 + tan2  + y. 2tan 1 + tan2  = x. 1 – y 2 x 2 1 + y 2 x 2 + y. 2. y x 1 + y 2 x 2 = x. x 2 – y 2 x 2 + y2 + 2xy2 x 2 + y2 = x 3 – xy 2 x 2 + y2 + 2xy2 x 2 + y2 = x 3 – xy 2 + 2xy2 x 2 + y2 = x 3 + xy2 x 2 + y2 = x(x 2 + y2 ) x 2 + y2 = x = R.H.S (Showed) (L) GLv‡b, OA = a = 12 km OB = b = 9 km AB = c = ? ⸪ ON || BN  BOT = OBN = 45  AOB = 180 – BOT = 180 – 45 = 135  C = AOB = 135 B A x 12 km 45 N 9 km O 45 T 135 N Avgiv Rvwb, cosC = a 2 + b2 – c 2 2ab  cos135 = 144 + 81 – c 2 216  – 1 2 = 225 – c 2 216  – 216 2 = 225 – c 2  c 2 = 225 + 216 2  c 2 = 377.73506  c = 19.4354 km  AB = 19.44 km (Ans.) (M) †`Iqv Av‡Q, T = l + m + n Q l R P T = l + m + n n m GLb, 1 T – m + 1 T – l = 3 T  1 l + n + 1 m + n = 3 l + m + n  m + n + n + l lm + nl + mn + n2 = 3 l + m + n  ml + m2 + mn + 2nl + 2mn + 2n2 + l 2 + ml + nl = 3lm + 3nl + 3mn + 3n2  m 2 + l 2 – n 2 = ml  m 2 + l 2 – n 2 ml = 1  m 2 + l 2 – n 2 2ml = 1 2  cosR = cos60  R = 60 (Ans.) 4| PQR GKwU wÎfzR| [h‡kvi †evW©- Õ23] (K) cÖgvY Ki †h, 2 cosx = 2 + 2 + 2 cos4x (L) DÏxcK n‡Z cÖgvY Ki †h, 1 + 4 sinQ + R 4 . sin R + P 4 . sinP + Q 4 = sinP 2 + sinQ 2 + sinR 2 (M) DÏxcK n‡Z cÖgvY Ki †h, p 3 cos(Q – R) + q3 cos(R – P) + r3 cos(P – Q) = 3pqr mgvavb: (K) R.H.S = 2 + 2 + 2 cos4x = 2 + 2(1 + cos4x) = 2 + 2.2cos2 2x = 2 + 2 cos2x = 2(1 + cos2x) = 2.2 cos2 x = 2 cosx = L.H.S (Proved) (L) PQR wÎfz‡R, P + Q + R  R.H.S = sinP 2 + sinQ 2 + sinR 2 =     sinP 2 + sinQ 2 + sinR 2 + 1 – 1 = 1 + 2 sinP + Q 4 .cos P – Q 4 + sinR 2 – sin 2 = 1 + 2 sinP + Q 4 .cos P – Q 4 + 2cos R +  4 .sinR –  4 = 1 + 2 sinP + Q 4 .cos P – Q 4 + 2cos 2R + P + Q 4 .sin– P – Q 4 [⸪ P + Q + R = ] = 1 + 2.sinP + Q 4 .cos P – Q 4 – 2cos P + Q + 2R 4 .sinP + Q 4 = 1 + 2sinP + Q 4     cos P – Q 4 – cos P + Q + 2R 4
4  Higher Math 1st Paper Chapter-7 = 1 + 2sinP + Q 4 . 2sinP – Q + P + Q + 2R 4  2 . sinP + Q + 2R – P + Q 4  2 = 1 + 4sinP + Q 4 . sinP + R 4 . sinQ + R 4 = 1 + 4.sinP + Q 4 . sinR + P 4 . sinP + Q 4 = L.H.S (Proved) (M) PQR wÎfz‡R, p sinP = q sinQ = r sinR psinQ = qsinP psinR = rsinP Ges p = q cosR + r cosQ q = r cosP + p cosR r = p cosQ + q cosP GLb, P 3 cos(Q – R) = p{p2 cos(Q – R)} = p(p2 cosQ. cosR + p2 sinQ.sinR) = p(p cosQ.p cosR + p sinQ. p sinR) = p{(pcosQ + qcosP) (pcosR + rcosP – rcosP)} + qsinP rsinP = p {(r – q cosP) (q – r cosP) + q sinP.r sinP} = p(qr – q 2 cosP – r 2 cosP + qr cos2 P + qr sin2 P) = p{qr – (q2 + r2 ) cosP + qr (sin2 P + cos2 P)} = p{qr – (q2 + r2 ) cosP + qr.1} = p {2qr – (q2 + r2 ) cosP} = 2pqr – P(q2 + r2 ) cosP Abyiƒcfv‡e, q 3 cos(R – P) = 2pqr – q(r2 + p2 ).cosQ Ges r 3 cos(P – Q) = 2pqr – r(p2 + q2 ) cosR L.H.S = p3 cos(Q – R) + q3 cos(R – P) + r3 cos(P – Q) = 2pqr – p(q2 + r2 ) cosp + 2pqr – q(r2 + p2 ) cosQ + 2pqr – r(p2 + q2 ) cosR = 6pqr – pq2 cosp – r 2 p cosp – qr2 cosQ – p 2 q cosQ – rp2 cosR – q 2 r cosR = 6pqr – pq(p cos Q + q cosP) – qr(q cos R + r cosQ) – rp (r cosP + p cosR) = 6pqr – pq.r – qr.p – rp.q = 6pqr – 3pqr = 3pqr = R.H.S (Proved) 5| S T E r q p [Kzwgjøv †evW©- Õ23] (K) tan42 tan78 cot6 cot66 Gi gvb wbY©q Ki| (L) cÖgvY Ki †h, tan E 2 = p – q p + q cot    S – T 2 (M) hw` p 4 + q4 + r4 = 2p2 (q2 + r2 ) nq, Z‡e †`LvI †h, S = 45 A_ev 135 mgvavb: (K) tan42 tan78 cot6 cot66 = 2 sin78.sin42 2 cos78.cos42 . 2 sin66.sin6 2 cos66.cos6 = cos(78 – 42) – cos(78 + 42) cos(78 + 42) + cos(78 – 42) . cos(66 – 6) – cos(66 + 6) cos(66 + 6) + cos(66 – 6) = cos36 – cos120 cos120 + cos36 . cos60 – cos72 cos72 + cos60 = 5 + 1 4 + 1 2 – 1 2 + 5 + 1 4 . 1 2 – 5 – 1 4 5 – 1 4 + 1 2 = 3 + 5 5 – 1 . 3 – 5 5 + 1 = 9 – 5 5 – 1 = 4 4 = 1 (Ans.) (L) S T E r q p S + T + E =  p sinS = q sinT = r sinE = 2R R.H.S = p – q p + q . cot S – T 2 = 2R sinS – 2R sinT 2R sinS + 2R sinT . cot S – T 2 = sinS – sinT sinS + sinT . cot S – T 2 = 2cos S + T 2 . sinS – T 2 2sinS + T 2 . cos S – T 2 . cot S – T 2 = cot S + T 2 . tan S – T 2 . cot S – T 2 = cot  – E 2 . 1 cot S – T 2 . cot S – T 2 = cot     2 – E 2 = tan E 2 = L.H.S (Proved) (M) †`Iqv Av‡Q, p 4 + q4 + r4 = 2p2 (q2 + r2 )  p 4 + q4 + r4 – 2p2 q 2 – 2p2 r 2 = 0  p 4 + q4 + r4 + 2r2 q 2 – 2p2 q 2 – 2p2 r 2 = 2r2 q 2  (r2 + q2 – p 2 ) = 2r2 q 2  r 2 + q2 – p 2 =  2rq  r 2 + q2 – p 2 2rq =  2rq 2rq