Tiêu đề TRIGONOMETRIC FUNCTIONS A-7.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :7 821 (a) Given, sin 2x + cos 4x = 2 ⇒ sin 2x + 1 ― 2 sin2 2x = 2 ⇒ 2 sin2 2x ― sin 2x + 1 = 0 Now, Discriminant, D = ( ―1) 2 ―4.2.1 = ―7 < 0 Hence, it is an imaginary equation, so the real root does not exist. 822 (d) We have, sin θ1 + sin θ2 + sin θ3 = 3 ⇒ sin θ1 = sin θ2 = sin θ3 = 1 [ ∵ ―1 ≤ sin x ≤ 1] ⇒θ1 = θ2 = θ3 = π 2 ⇒ cos θ1 + cos θ2 + cos θ3 = 0 823 (b) We have, k sin x + (1 ― 2 sin2 x) = 2k ― 7 ⇒2 sin2 x ― k sin x + 2(k ― 4) = 0 ⇒ sin x = k ± k 2 ― 16 k + 64 4 = k ± (k ― 8) 4 = 1 2 (k ― 4),2 ⇒ sin x = 1 2 (k ― 4) [ ∵ sin x ≠ 2 ] Now, ―1 ≤ sin x ≤ 1⇒ ― 1 ≤ k ― 4 2 ≤ 1⇒2 ≤ k ≤ 6 824 (b) Given that, cos 2B = cos(A + C) cos(A ― C) = cosA cos C ― sinA sin C cosA cos C + sinA sin C ⇒ 1 ― tan2B 1 + tan2B = 1 ― tanA tan C 1 + tanA tan C ⇒1 + tan2B ― tanA tan C ― tanA tan C tan2B Topic :-TRIGONOMETRIC FUNCTIONS Solutions
= 1 ― tan2B + tanA tan C ― tanA tan C tan2B ⇒2 tan2B = 2 tanA tan C ⇒ tan2B = tanA tan C Hence, tanA, tanB and tan C will be in GP 825 (c) We have, ( cosA + cosB sinA ― sinB ) n + ( sinA + sinB cosA ― cosB ) n = (cot A ― B 2 ) n + ( ― cot A ― B 2 ) n = {1 + ( ―1) n }cotn ( A ― B 2 ) = 0 × cotn ( A ― B 2 ) = 0 [ ∵ n is odd] 826 (a) We have, a tan θ + b sec θ = c ⇒b sec θ = c ― a tan θ ⇒b 2 sec2 θ = c 2 + a 2 tan2 θ ― 2 ac tan θ ⇒b 2 (1 + tan2 θ) = c 2 + a 2 tan2 θ ― 2 ac tan θ ⇒ tan2 θ(b 2 ― a 2 ) + 2 ac tan θ + b 2 ― c 2 = 0 Since tanα and tanβ are roots of this equation ∴ tanα + tanβ = ―2 ac b 2 ― a 2 and tanα tanβ = b 2 ― c 2 a 2 ― c 2 Now, tan(α + β) = tanα + tanβ 1 ― tanα tanβ = ― 2 ac b 2 ― a 2 1 ― b 2 ― c 2 b 2 ― a 2 = 2 ac a 2 ― c 2 827 (b) tan θ + tan θ + 3 1 ― 3tan θ + tan θ ― 3 1 + 3tan θ = 3 ⇒ tan θ + 8 tan θ 1 ― 3 tan2 θ = 3 ⇒ 9 tan θ ― 3 tan3 θ 1 ― 3 tan2 θ = 3 ⇒ 3 tan 3θ = 3 ⇒ tan 3θ = 1 828 (a)
Since, y = 1 + 4sin2 x cos2 x ⇒y = 1 + sin2 2x We know that, 0 ≤ sin2 2x ≤ 1 ⇒1 ≤ 1 + sin2 2x ≤ 2 ⇒1 ≤ y ≤ 2 829 (a) sin2α + sin2β ― sin2 γ = sin2α + sin(β ― γ) sin(β + γ) = sin2α sin(π ― α)sin(β + γ) [ ∵ α + β ― γ = π] = sinα[ sinα + sin(β + γ) ] = sinα[ sin{π ― (β ― γ)} + sin(β + γ) ] = sinα[ sin(β ― γ) + sin(β + γ) ] = sinα[2 sinβ cos γ ] = 2 sinα sinβ cos γ 830 (a) We have, cot A 2 + cot B 2 + cot C 2 = s(s ― a) ∆ + s(s ― b) ∆ + s(s ― c) ∆ = s ∆ (3s ― 2s) = s 2 ∆ And, cotA + cotB + cot C = cosA sinA + cosB sinB + cos C sin C = b 2 + c 2 ― a 2 2 bc sinA + c 2 + a 2 ― b 2 2ac sinB + a 2 + b 2 ― c 2 2ab sin C = b 2 + c 2 ― a 2 4 ∆ + c 2 + a 2 ― b 2 4 ∆ + a 2 + b 2 ― c 2 4 ∆ = a 2 + b 2 + c 2 4 ∆ ∴ cot A 2 + cot B 2 + cot C 2 cotA + cotB + cot C = s 2 ∆ a 2 + b 2 + c 2 4 ∆ = (2s) 2 a 2 + b 2 + c 2 = (a + b + c) 2 a 2 + b 2 + c 2 831 (a) ∵ (tanα ― cot α) 2 ≥ 0
⇒ tan2α + cot2α ― 2 ≥ 0 ⇒ tan2α + cot2α ≥ 2 832 (a) We have, tan m θ = tan n θ ⇒m θ = t π + n θ, where r ∈ Z ⇒θ = r π m ― n ,r ∈ Z Clearly, these values of θ from an A.P. with common difference π m ― n 833 (a) We have, sinA sin C = sin(A ― B) sin(B ― C) ⇒ sin(B + C) sin(a + B) = sin(A ― B) sin(B ― C) ⇒ sin2B ― sin2 C = sin2A ― sin2B ⇒b 2 ― c 2 = a 2 ― b 2 ⇒a 2 ,b 2 ,c 2 are in A.P. 834 (a) Let sec θ ― tan θ = λ ...(i) Then, (sec θ + tan θ) = 1 sec θ ― tan θ ⇒sec θ + tan θ = 1 λ ...(ii) ∴ 2tan θ = 1 λ +λ [On subtracting (i) from (ii)] ⇒2x ― 1 2x = 1 λ ― λ ⇒λ = 1 2x , ― 2x⇒ sec θ ― tan θ = 1 2x , ― 2x 835 (d) We observe that the LHS of the given equation is not defined for x = n π,n ∈ Z Now, cot x ― cosec x = 2 sin x ⇒ cot x ― 1 = 2 sin2 x ⇒2 cos2 x + cos x ― 3 = 0 ⇒(2 cos x + 3)(cos x ― 1) = 0 ⇒cos x = 1 [ ∵ 2 cos x + 3 ≠ 0] ⇒x = 0, 2 π But, x ≠ n π,n ∈ Z Hence, the given equation has no solution 837 (d)