CG SCERT Chhattisgarh State Board Class 10 Mathematics Chapter 9 Trigonometrical equations and Identities Exercises Questions and Answers.
Exercise – 1
LHS;
(1) 1/Sec θ – 1 – 1/Sec θ + 1 = (Sec θ + 1) – (Sec θ – 1)/Sec2 θ – 12 = 2/tan2 θ = 2oot = θ = RHS
(2)
LHS;
Sec2θ. Cosec2θ = (1 + tan2θ) (1 + cot2θ) = tan2 θ + 1 + 1 + cot2 θ = 2 + tan2 θ + cot2 θ
RHS,
Sec2θ. Cosec2θ = (1 + tan2θ) (1 + cot2θ) = 1 + Cot2θ + tan2θ + tan2θ . cot2θ.
= 1 + 1 + cot2 θ + tan2 θ
= 2 + tan2 θ + cot2 θ
∴ LHS = RHS
(3) Sin4A + Cos4A = 1 – 2 sin2A . Cos2 A
Taking; Sin2 A + Cos2 A = 1
= Sin4A + Cos4A + 2 Sin2A Cos 2 A = 12 [sq. both sides]
= Sin4 A + Cos4 A = l – 2sin2 A cos2 A [Proved]
(4)
√1 – Cos θ/ 1 + Cos θ = Cosec θ – Cot θ
LHS, √1 – Cos θ/1 + Cos θ = √(1 – Cos θ) × (1 – Cos θ)/ (1 + Cos θ) × (1 – Cos θ)
= √(1 – Cos θ)2/1 – Cos2θ = 1 – Cos θ/Sinθ = Cosec θ – Cot θ = RHS [Proved]
(5) (1 + Cot θ – Cosec θ) (l + tan θ + Sec θ) = 2
LHS, (1 + Cot θ – Cosec θ) (1 + tan θ + Sec θ)
= (1 + Cos θ/Sin θ -1/Sin θ) (1 + Sin θ/Cos θ + 1/Cos θ)
= [Sin θ + Cos θ – 1/Sin θ) × [Cos θ + Sin θ + 1/Cos θ] = (Sin θ + Cos θ – 1) (Sin θ + Cos θ + 1/Sin θ Cos θ)
= (Sin θ + Cos θ)2 – 12/Sin θ Cos θ = [Sin2 θ + Cos2 θ + 2 Sin θ Cos θ/ -1]/ Sin θ Cos θ
= 1 + 2Sin θ Cos θ – 1/Sin θ Cos θ
= 2 Sin θ Cos θ / Sin θ Cos θ
= 2 = RHS [ Proved]
(6) 1 + Cos θ/1 – Cos θ – 1 – Cos θ/1 + Cos θ = 4 Cot θ Cosec θ
LHS, 1 + Cos θ/1 – Cos θ – 1 – Cos θ/1 + Cos θ = (1 + Cos θ)2 – (1 – Cos θ)2/12 – Cos2 θ = 2 Cos θ/Sin2 θ = 2 Cot θ Cosec θ = RHS
(7) Sin θ/1 + Cos θ + 1 + Cos θ/Sin θ = 4Cot θ Cosec θ
LHS, Sin2 θ + (Ncosθ)2/Sin θ (1 + Cos θ) = [Sin2 θ + Cos2 θ + 1 + 2 Cos θ/ Sin θ (1 + Cos θ) = 2 (1 + Cos θ)/Sin θ (1 + Cos θ) = 2 Cosec θ – RHS.
(8) Cos θ – Sin θ = √2 Sin θ
= (Cos θ – Sin θ)2 = 2Sin2 θ [Sq. both Side]
= Cos2 θ – 2Cos θ Sin θ = Sin2θ
= 2Cos2 θ – 2Cos θ Sin θ = Sin2 θ + Cos2 θ [adding Cos2 θ both sides]
= 2Cos2 θ = (Sin θ + Cos θ)2
= Sin θ + Cos θ = √2 Cos θ [Proved]
(9) tan θ = n tan and Sin θ = m Sin θ Prove Cos2 θ = m2 – 1/n2 -1
Now, Sin θ = m sin θ – (i)
Tan θ = n tan θ = Sin θ/Cos θ = n sin θ/Cos θ – (ii)
From (i) and (ii); Cos θ = n/m Cos θ – (iii)
Sin2 θ = m2 SPA2
= (1 – Cos2 θ) = m2 (1 – Cos2) = 1 – Cos2 θ = m2 [1 – n2/m2 Cos2]
= Cos2 θ = m2 – 1/n2 – l (Proved)
(10) x = a Cosec θ = x/a = Cosec θ – (i)
- b Cot θ = y/b= Cot θ – (ii)
S.Y. the equations and subtraction (ii from i)
x2/a2 – y2/b2 = Cosec2 θ – Cot2 θ = 1 [1 + Cot2 θ = Cosec2 θ]
(11) x2 + y2 + z2 = r2 Sin2 A Cos2C + r2 Sin2A Sin2C + r2Cos2 A
= r2 Sin2 A (Cos2 C + Sin2 C) + r2 Cos2 A
= r2 Sin2 A + r2 Cos2 A
= r2
Exercise 2:
(i) 2Cos2 θ – √3cos θ ⇒cos θ (2cos θ – √3) = 0
∴cos θ = √3/2
∴ 0 = 30
(ii) 2Sin θ – Cos θ = 1
= 2 – 2Cos2 θ – Cos θ = 1
= Cos θ (2Cos θ + 1) = 1
∴θ = 0°
(iii) 3 tan2θ = 2 Sec2 θ + 1
= tan2θ = 2sec2θ – 2 tan2θ + 1
= tan2 θ = 3
= tan θ = √3
= θ = 60°
(iv) Cos2θ – 3 Cos θ + 2 = Sin2 θ
= Cos2 θ – 3Cos θ + 2 = 1 – Cos2 θ
= 2 Cos2 θ – 2Cos θ – Cos θ + 1 = 0
= 2Cos θ (Cos θ – 1) -1 (Cos θ – 1) = 0
= Cos θ = 1/2 or Cos θ = 1
= θ = 60° or 0 = 0°
(v) Cos θ/1 – Sin θ + Cos θ/1 + Sin θ = 4
= Cos θ [(4sin θ) + (1 – sin θ)]/ 1 – in2 θ = 2Cos θ/Cos2 θ = 4
= 2/Cos θ = 4 = Cos θ = 1/2
= θ = π/3 = 60°
Exercise – 3
(1) (i) Sin56° = Sin (90 – 34) = Cos 34°
(ii) tan 81° = tan (90 – 9) =Cot 9°
(iii) Sec 73° = sec (90 – 17) = cosec 17°
(2) (i) cos 80°/sin10° = cos (90 – 10)/sin 10°
= sin10/sin10 = 1
(ii) Sin 37°/2cos 53° = Sin (90 – 53)/2cos 53
= Cos 53/2Cos 53
= 1/2
(iii) 3 Sin 17 Sec 73°
= 3 Sin (90 – 73) Sec 73°
= 3 Cos 73° Sec 73°
= 3
(3) (i) Sin 64° – Cos 26° = Sin (90 – 26) – Cos 26°
= Cos 26° – Cos 26°
= 0
(ii) 3 Cos 80° Cosec 10° + 2 Cos 59° Cosec 31°
= 3 Sin 10 Cosec 31°
= 3 + 2
= 5.
(iii) 2 Cos 67°/Sin 23° – tan 40°/Cot 50° + Cos 0°
= 2 Sin 23°/Sin 23 – Cot 50°/Cot 50 + 1
= 2 – 1 + 1
= 2
(iv) Sin235 + Sin2 55 = Sin2 35 + Cos2 35 = 1
(v) [5 sin 35°/cos 55°] + cos 55°/2son 35°] – 2Cos 60°
= 5 Cos 55°/Cos 55° + Sin 35°/2Sin 35° – 2 Cos 60°
= 5 + 1/2 – 2. 1/2
= 5 + 1/2 – 1
= 7/2
(4) (i) Sin 63° Cos 27° + Cos 63° Sin 27° = 1
RHS; Sin 63° Cos (90 – 63) + Cos 63° Sin (90 – 63°)
= Sin2 63° + Cos2 63
= 1. RHS
(ii) tan 15° tan 36° tan 45° tan 54° tan 75° = 1
LHS; Cot 75° tan 75° tan 45° tan 36° Cot 36°
= 1/tan 75° × tan 45° × tan 36° × 1/tan 36°
= 1 = RHS
(iii) Sin285 + Sin280 + Sin210 + Sin25 = 2
LHS; Sin2 85 + Cos2 5 + Sin2 80° + Cos2 80° = 1.RHS.
(5) Sin (90° – θ) Cos (90° – θ) = tan θ/1 + Cot2 (90° – θ)
LHS, Sin (90° – θ) Cos (90° – θ)
= Cos θ Sin θ
RHS, tan θ/1 + Cot2 (90 – θ) = tan θ/1 + tan2 θ = tan θ/Sec2 θ = Sin θ/Cos θ × Cos2θ
= Sin θ Cos θ
∴LHS = RHS
(6) Cos θ/Sec (90° – θ) + 1 + Sin (90° – θ)/Cosec θ – 1 = 2 Cot (90° – θ)
LHS, Cos θ/Sec (90° – θ) + 1 + Sin (90° – θ)/Cosec θ – 1 = Cos θ/Cosec θ 1 + Cos θ/Cosec θ -1 = 2Cot θ/Cosec2 θ – 1 = 2 Cos θ/Sin θ × Sin2θ/Cos2 θ = 2 tan θ
RHS, 2Cot (90° – θ) = 2 tan θ
∴LHS = RHS
(7) tan (90° – θ)/Cosec2 θ tan θ = Cos2 θ
LHS tan (90° – θ)/Cosec2 θ tan θ = Cot θ/Cosec2 θ tan θ
= Cos θ/Sin θ × Sin2 θ × Cos θ/Sin θ
= Cos2 θ = LHS
(8) Sin A = Cos B
= Sin A = Sin (90° – B) = A = 90 – B
= A + B = 90
(9) Cosec 2A = Sec (A – 36°)
= Sec (90° – 2A) = Sec (A – 36°)
= 90° – 2A = A – 36
= 3A = 126
= A = 42°
(10) A + B = 90° = A = 90° – B
a2 – b2 = Cosec2B – Cot2B
= 1
Sec A = a = Sec (90° – B) = a = a = Cosec B
Cot B = b ⇒
(11) A + B + C = 180° = B + C = 180° – A
⇒ B + C/2 = 90° – A/2
⇒tan [B + C/2] tan [90° – A/2] = 0
(12) Cot2 56° + Cosec 56 = Cot2 (90 – 34) + Cosec (90 – 34)
= tan2 34 + Sec 34
= [(Sec2 = (34) -1)] = Sec 34
= x2 + x – 1