Title TRIGONOMETRIC FUNCTIONS A-9.pdf ✅

Class : XIth Subject : MATHS Date : DPP No. :9 861 (c) cosec 15° + sec 15° = 2(sin 15° + cos 15°) 2 sin 15° cos 15° = 2[ 3 2 2 ― 1 2 2 + 3 2 2 + 1 2 2 ]/ sin 30° = 4 3 2 = 2 6 862 (d) We have, sinA = 4 5 andcosB = ― 12 13 Now,cos(A + B) = cosA cosB ― sinA sinB = 1 ― 16 25 ( ― 12 13) ― 4 5 1 ― 144 169 = ― 3 5 × 12 13 ― 4 5 ( ― 5 13) = ― 36 65 + 20 65 = ― 16 65 863 (a) Given that, 1 tan θ + tan θ = m ⇒1 + tan2 θ = m tan θ ⇒sec2 θ = m tan θ ...(i) and sec θ ― cos θ = n ⇒ sec2 θ ― 1 = n sec θ ⇒ tan2 θ = n sec θ ⇒tan4 θ = n 2 sec2 θ = n 2 ∙ m tan θ [from Eq.(i)] ⇒ tan3 θ = n 2m ( ∵ tan θ ≠ 0) ⇒tan θ = (n 2m) 1/3 ...(ii) Topic :-TRIGONOMETRIC FUNCTIONS Solutions
From Eq. (i), we get sec2 θ = m (n 2m) 1/3 As we know that, sec2 θ ― tan2 θ = 1 ⇒m(mn 2 ) 1/3 ― (n 2m) 2/3 = 1 ⇒m(mn 2 ) 1/3 ― n(nm2 ) 1/3 = 1 864 (c) We have, (sinA + sinB + sin C)(sinA + sinB ― sin C) = 3 sinA sinB ⇒(sinA + sinB) 2 ― sin2 C = 3 sinA sinB ⇒ sin2A + sin2B ― sin2 C = sinA sinB ⇒ sin2A + sin(B + C)sin(B ― C) = sinA sinB ⇒ sin2A + sinA sin(B ― C) = sinA sinB ⇒ sinA[ sin(B + C) + sin(B ― C) ] = sinA sinB ⇒2 sinA sinB cos C = sinA sinB ⇒ cos C = 1/2 [ ∵ sinA sinB ≠ 0 ] ⇒C = 60° 865 (b) Given, cos 2x + 7 = a (2 ― sin x) ⇒ 1 ― 2 sin2 x + 7 = 2a ― a sin x ⇒ 2 sin2 x ― a sin x + (2a ― 8) = 0 ∴ sin x = a ± ( ―a) 2 ― 8(2a ― 8) 2 × 2 = a ± (a ― 8) 4 For ( + ) sign, sin x = a ― 4 2 For ( ― ) sign, sin x = 2 which is not possible We know ―1 ≤ sin x ≤ 1 ∴ ― 1 ≤ a ― 4 2 ≤ 1 ⇒ 2 ≤ a ≤ 6 866 (c) cos2B + cos2 C = cos2B + cos2 ( π 2 ― B) = cos2B + sin2B = 1 867 (d) We have,
b 2 sin 2 C + c 2 sin 2 B = b 2 ∙ (2 sin C cos C) + c 2 ∙ (2 sinB cosB ) = 2(b sin C)(b cos C) + 2(c sinB )(c cosB ) = 2(c sinB)(b cos c) + 2(c sinB )(c cosB ) [ ∵ b sinB = c sin C ] = 2 c sinB(b cos C + c cosB ) = 2 ac sinB = 4 ∆ 868 (a) Since the angles of ∆ABC are in A.P. ∴ 2B = A + C⇒3B = A + B + C⇒3B = 180°⇒B = 60° Now, sinA a = sinB b ⇒ sinA = a b sinB = 24 22 sin 60° = 6 3 11 ⇒ cosA = 13 11 We have, sin C = sin{180° ― (A + B)} ⇒ sin C = sin(A + B) ⇒ sin C = sinA cosB + cosA sinB ⇒ sin C = ( 6 3 11 )( 1 2 ) + 13 11 ( 3 2 ) = 6 3 + 39 22 ∴ c = b sin C sinB ⇒c = 12 + 2 13 869 (a) We have, ∠BFC = π 2 = ∠BEC So, the circle with BC as diameter will pass through E and F. Clearly, the circle with BC as diameter is the circumcircle of ∆BEF such that ∠FBE = 90° ― A ∴ FE = 2( a 2 ) sin∠FBE [Using :a = 2R sinA ] ⇒FE = a cosA Let R1 be the radius of the circumcircle of ∆DEF. Then, R1 = FE 2 sin∠FDE = a cosA 2 sin(180° ― 2A) ⇒R1 = a cosA 4 sinA cosA = a 4 sinA = R 2
870 (b) Clearly, the equation x 2 + 2x + 1 = 0 has imaginary roots. So, the two equations have both common roots ∴ a 1 = b 2 = c 1 ⇒ sinA 1 = sinB 2 = sin C 1 ⇒ sinA 1/ 2 = sinB 1 = sin C 1/ 2 ⇒A = π 4 ,B = π 2 ,C = π 4 871 (a) We have, s = 3a 2 and ∆ = 3 4 a 2 ∴ r = ∆ s = a 2 3 Let the length of each side of the square inscribed in the incircle be x. Then, x 2 + x 2 = (Diameter) 2 ⇒2x 2 = a 2 3 ⇒x 2 = a 2 6 ⇒Area of the square = a 2 6 872 (a) cos1°. cos 2° ..... cos 179° = cos1°. cos 2° .... cos 90° . cos 179° = 0 [ ∵ cos 90° = 0] 873 (a) Given equation is 2 cos 2x + 1 = 3.2— sin x By taking option (a) Let x = nπ When, n = 1, x = π ∴ 2 cos 2π + 1 = 3.2―sin π ⇒ 2 + 1 = 3.2° ⇒ 3 = 3 When n = 2, x = 2π ∴ 2 cos 4π + 1 = 3.2 ―sin 2π ⇒ 2 1 + 1 = 3.2° ⇒ 3 = 3 A B D C E F O