Title Inverse Trigonometric Practice Sheet Solution HSC FRB 25.pdf ✅

2  Higher Math 2nd Paper Chapter-7  cos    x +  4 = – 1 2  cos    x +  4 = cos 3 4  x +  4 = 2n  3 4 ; n  Z  x = 2n  3 4 –  4  x = 2n + 3 4 –  4 ; 2n – 3 4 –  4 = 2n +  2 , 2n –  = (4n + 1) 2 , 2n –  n = 0 n‡j, x =  2 , –  n = 1 n‡j, x = 5 2 ,  n = – 1 n‡j, x = – 3 2 , – 3 – 2 < x < 2 e ̈ewa‡Z, x = – 3 2 , – ,  2 ,  wb‡Y©q mgvavb: x = – 3 2 , – ,  2 ,  (Ans.) 2| f(x) = sin–1 p + sin–1 q + sin–1 r, A = cosx – cos2x, R = 1 – cosx [ivRkvnx †evW©- Õ23] (K) cÖgvY Ki †h, tan–11 3 = 1 2 sin–13 5 (L) f(x) =  n‡j †`LvI †h, p 1 – p 2 + q 1 – q 2 + r 1 – r 2 = 2pqr (M) mgvavb Ki: A R = 1 ; hLb 0 < x <  mgvavb: (K) L.H.S = tan–11 3 = 1 2  2tan–11 3 = 1 2 sin–1 2. 1 3 1 +     1 3 2 = 1 2 sin–1 2 3 1 + 1 9 = 1 2 sin–1 2 3 10 9 = 1 2 sin–1     2 3  9 10 = 1 2 sin–13 5 = R.H.S (Proved) (L) †`Iqv Av‡Q, f(x) = sin–1 p + sin–1 q + sin–1 r awi, sin–1 p = A, sin–1 q = B Ges sin–1 r = C  sinA = p  sinB = q  sinC = r GLb, f(x) =  n‡j, sin–1 p + sin–1 q + sin–1 r =   A + B + C =  L.H.S = p 1 – p 2 + q 1 – q 2 + r 1 – r 2 = sinA 1 – sin2A + sinB 1 – sin2B + sinC 1 – sin2C = sinA cosA + sinB cosB + sinC cosC = 1 2 (2sinA cosA + 2sinB cosB + 2sinC cosC) = 1 2 (sin2A + sin2B + sin2C) = 1 2     2sin2A + 2B 2 cos 2A – 2B 2 + 2sinC cosC = 1 2 [2sin(A + B) cos(A – B) + 2sinC cosC] = sin( – C) cos(A – B) + sinC cosC [⸪ A + B + C = ] = sinC cos(A – B) + sinC cosC = sinC [cos(A + B) + cos{ – (A + B)}] [⸪ A + B + C = ] = sinC [cos(A – B) – cos(A + B)] = sinC . 2sinA sinB = 2sinA sinB sinC = 2pqr = R.H.S (Showed) (M) †`Iqv Av‡Q, A = cosx – cos2x R = 1 – cosx GLv‡b, A R = 1 ; hLb 0 < x <   cosx – cos2x 1 – cosx = 1  cosx – 2cos2 x + 1 = 1 – cosx  2cos2 x – cosx – 1 = cosx – 1  2cos2 x – 2cosx = 0  2cosx (cosx – 1) = 0  cosx (cosx – 1) = 0 nq, cosx = 0  x = (2n + 1) 2 ; n  Z A_ev, cosx = 1  x = 2n ; n  Z n = 0 n‡j, x =  2 , 0 n = 1 n‡j, x = 3 2 , 2 ; Bnv MÖnY‡hvM ̈ bq| †Kbbv 0 < x <   0 < x <  e ̈ewa‡Z x =  2 (Ans.) 3| sinA = 2 5 , cosB = 4 5 , cotC = 3 Ges g() = cos – cos7 [h‡kvi †evW©- Õ23] (K) cÖgvY Ki †h, sin.tan–1 .cot. cos–1 y = y. (L) cÖgvY Ki †h, A – 1 2 B + C = tan–1 2. (M) hw` g() = sin4 nq, Zvn‡j  Gi gvb wbY©q Ki|

4  Higher Math 2nd Paper Chapter-7 mgvavb: (K) †`Iqv Av‡Q, sin–1m + sin–1 n =  2  sin–1m =  2 – sin–1 n  m = sin     2 – sin–1 n  m = cos(sin–1 n)  m 2 = cos2 (sin–1 n)  m 2 = 1 – sin2 (sin–1 n)  m 2 = 1 – {sin(sin–1 n)}2  m 2 = 1 – n 2  m 2 + n2 = 1 (Showed) (L) †`Iqv Av‡Q, x = a cosP  cosP = x a  P = cos–1x a Ges y = b cosQ  cosQ = y b  Q = cos–1y b GLv‡b, P + Q =   cos–1x a + cos–1y b =   cos–1      x  a . y b –     1 – x 2 a 2     1 – y 2 b 2 =   xy ab –     1 – x 2 a 2     1 – y 2 b 2 = cos  –     1 – x 2 a 2     1 – y 2 b 2 = cos – xy ab      1 – x 2 a 2     1 – y 2 b 2 = cos2 – 2 xy ab cos + x 2 y 2 a 2 b 2  1 – x 2 a 2 – y 2 b 2 + x 2 y 2 a 2 b 2 = cos2 –2 xy ab cos + x 2 y 2 a 2 b 2  – x 2 a 2 – y 2 b 2 + x 2 y 2 a 2 b 2 – x 2 y 2 a 2 b 2 + 2 xy ab cos = cos2 – 1  –     x 2 a 2 – 2 xy ab cos + y 2 b 2 = – (1 – cos2)  x 2 a 2 – 2 xy ab cos + y 2 b 2 = sin2 (Proved) (M) †`Iqv Av‡Q, f(z) = tanz . tan3z GLb, f(z) = 1 n‡j, tanz . tan3z = 1  sinz sin3z cosz cos3z = 1  cos3z cosz = sin3z sinz  cos3z cosz – sin3z sinz = 0  cos(3z + z) = 0  cos4z = 0  4z = (2n + 1)  2 , n  Z  z = (2n + 1) 8 n = 0 n‡j, z =  8 n = – 1 n‡j, z = –  8 n = 1 n‡j, z = 3 8 n = – 2 n‡j, z = – 3 8 n = 2 n‡j, z = 5 8 ; Bnv MÖnY‡hvM ̈ bq, †Kbbv –  2 < z <  2 –  2 < z <  2 e ̈ewa‡Z, z = – 3 8 , –  8 ,  8 , 3 8 (Ans.) 5| f(x) = cosx – cos7x Ges g(x) = sinx [Kzwgjøv †evW©- Õ23] (K) cÖgvY Ki †h, sin–1 2x 1 + x2 = cos–11 – x 2 1 + x2 (L) f() = sin4 mgxKi‡Yi mvaviY mgvavb wbY©q Ki| (M) g     g      2 – x = g      2 – g(x) n‡j †`LvI †h, x =   4 + cos–1 1 2 2 . mgvavb: (K) awi, x = tan   = tan–1 x L.H.S = sin–1 2x 1 + x2 = sin–1 2 tan 1 + tan2  = sin–1 (sin2) = 2 = 2tan–1 x R.H.S = cos–1 – x 2 1 + x2 = cos–11 – tan2  1 + tan2  = cos–1 (cos2) = 2 = 2tan–1 x  L.H.S = R.H.S (Proved) (L) †`Iqv Av‡Q, f(x) = cosx – cos7x  f() = cos – cos7 GLv‡b, f() = sin4  cos – cos7 = sin4  2sin7 +  2 sin7 –  2 = sin4  2sin4 sin3 – sin4 = 0  sin4(2sin3 – 1) = 0 nq, sin4 = 0  4 = n ; n  Z   = n 4 A_ev, 2sin3 = 1  sin3 = 1 2  sin3 = sin 6  3 = n + (– 1)n 6 ; n  Z   = n 3 + (– 1)n  18 wb‡Y©q mgvavb:  = n 4 , n 3 + (– 1)n  18 , †hLv‡b n  Z (Ans.)