Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Solutions

(b) cos A Hint: L.H.S. = sin [(n + 1)A]. sin [(n + 2)A] + cos [(n + 1)A]. cos [(n + 2)A] = cos [(n + 2)A]. cos [(n + 1)A] + sin [(n + 2)A]. sin [(n + 1)A] Let (n + 2)A = a and (n + 1)A = b … (i) ∴ L.H.S. = cos a. cos b + sin a. sin b = cos (a – b) = cos [(n + 2)A – (n + 1)A] ……..[From (i)] = cos [(n + 2 – n – 1)A] = cos A = R.H.S.

(c) Hint:

(d) tan α Hint:

(c) Hint:

(c) cos 3A Hint:

(b) Hint:

(c) 2 sin α sin β cos γ Hint: sin2 α + sin2 β – sin2 γ = = 1 – (cos 2α + cos 2β) – 1 + cos2 γ = × 2 cos(α + β) cos(α – β) + cos2 γ = cos γ cos (α – β) + cos2 γ …..[∵ α + β + γ = π] = cos γ [cos (α – β) + cos γ] = cos γ [cos (α – β) – cos (α + β)] = 2 sin α sin β cos γ

(b) Hint: 3 sin 2A – 2sin 2B = 0 sin 2B = sin 2A …….(i) 3 sin2 A + 2 sin2 B = 1 3 sin2 A = 1 – 2 sin2 B 3 sin2 A = cos 2B ……(ii) cos(A + 2B) = cos A cos 2B – sin A sin 2B = cos A (3 sin2 A) – sin A ( sin 2A) …..[From (i) and (ii)] = 3 cos A sin2 A – (sin A) (2 sin A cos A) = 3 cos A sin2 A – 3 sin2 A cos A = 0 = cos ∴ A + 2B = ……..[∵ 0 < A + 2B < ]

(a) acute angled Hint: cot A cot B cot C > 0 Case I: cot A, cot B, cot C > 0 ∴ cot A > 0, cot B > 0, cot C > 0 ∴ 0 < A < , 0 < B < , 0 < C < ∴ ∆ABC is an acute angled triangle. Case II: Two of cot A, cot B, cot C < 0 0 < A, B, C < π and two of cot A, cot B, cot C < 0 ∴ Two angles A, B, C are in the 2nd quadrant which is not possible.

(a) √3 Hint: L.H.S. = tan 20° tan 80° cot 50° = tan 20° tan 80° cot (90° – 40°) = tan 20° tan 80° tan 40° = tan 20° tan (60° + 20°) tan (60° – 20°) = tan 3(20°) = tan 60° = √3 = R.H.S.

L.H.S. = tan 20° tan 80° cot 50° = tan 20° tan 80° cot (90° – 40°) = tan 20° tan 80° tan 40° = tan 20° tan (60° + 20°) tan (60° – 20°) = tan 3(20°) = tan 60° = √3 = R.H.S.

sin α sin β – cos α cos β + 1 = 0 ∴ cos α cos β – sin α sin β = 1 ∴ cos (α + β) = 1 ∴ α + β = 0 ……[∵ cos 0 = 1] ∴ β = -α L.H.S. = cot α tan β = cot α tan(-α) = -cot α tan α = -1 = R.H.S.

L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384° = cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°) = cosec 48° + cosec 84° + cosec (-12°) + cosec 24°

(sin x – cos x)4 = [(sin x – cos x)2]2 = (sin2 x + cos2 x – 2 sin x cos x)2 = (1 – 2 sin x cosx)2 = 1 – 4 sin x cos x + 4 sin2 x cos2 x (sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x sin6 x + cos6 x = (sin2 x)3 + (cos2 x)3 = (sin2 x + cos2 x)3 – 3 sin2 x cos2 x (sin2 x + cos2 x) …..[∵ a3 + b3 = (a + b)3 – 3ab(a + b)] = 13 – 3 sin2 x cos2 x (1) = 1 – 3 sin2 x cos2 x L.H.S. = 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) = 3(1 – 4 sin x cos x + 4 sin2 x cos2 x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin2 x cos2 x) = 3 – 12 sin x cos x + 12 sin2 x cos2 x + 6 + 12 sin x cos x + 4 – 12 sin2 x cos2 x = 13 = R.H.S.

We have to prove that, tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A i.e., to prove, cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0 ∴ cot θ – tan θ = 2 cot 2θ …..(i) L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)] = 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A = 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)] = 4(cot 4A – tan 4A) – 8 cot 8A = 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)] = 8 cot 8A – 8 cot 8A = 0 = R.H.S. Alternate Method:

sin A – cos B = cos C ∴ sin A = cos B + cos C A = B – C ………(i) In ∆ABC, A + B + C = π ∴ B – C + B + C = π ∴ 2B = π ∴ B =

L.H.S. = sin 20°. sin 40°. sin 80° = sin 20°. sin 40°. sin 80° = (2. sin 40°. sin 20°). sin 80° = [cos(40° – 20°) – cos (40° + 20°)]. sin 80° = (cos 20° – cos 60°) sin 80° = . cos 20°. sin 80° – . cos 60°. sin 80° = (2 sin 80°. cos 20°) – . sin 80° = [sin(80° + 20°) + sin (80° – 20°)] – . sin 80°

Let θ = 18° ∴ 5θ = 90° ∴ 2θ + 3θ = 90° ∴ 2θ = 90° – 3θ ∴ sin 2θ = sin (90° – 3θ) ∴ sin 2θ = cos 3θ ∴ 2 sin θ cos θ = 4 cos3 θ – 3 cos θ ∴ 2 sin θ = 4 cos2 θ – 3 …..[∵ cos θ ≠ 0] ∴ 2 sin θ = 4 (1 – sin2 θ) – 3 ∴ 2 sin θ = 1 – 4 sin2 θ ∴ 4 sin2 θ + 2 sin θ – 1 = 0 ∴ sin θ = = = Since, sin 18° > 0 ∴ sin 18°=

We know that, cos 2θ = 1 – 2 sin2 θ cos 36° = cos 2(18°) = 1 – 2 sin2 18° ∴ cos 36° =

We know that, sin2 θ = 1 – cos2 θ sin2 36° = 1 – cos2 36° = 1 – = = ∴ sin 36° = ……[∵ sin 36° is positive]