Maharashtra Board Class 11 Maths Solution Chapter 3 – Trigonometry 2
Balbharati Maharashtra Board Class 11 Maths Solution Chapter 3: Trigonometry 2. Marathi or English Medium Students of Class 11 get here Trigonometry 2 full Exercise Solution.
Std |
Maharashtra Class 11 |
Subject |
Maths Solution |
Chapter |
Trigonometry 2 |
Exercise – 3.1
(1) Find the values of
(i) sin 15°
(ii) cos 75°
(iii) tan 105°
(iv) cot 225°
Solution:
(i) sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
= 1/√2 . √3/2 – 1/√2 . 1/2
= √3 – 1/2√2
(ii) Cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°
= 1/√2 . √3/2 – 1/√2 . 1/2
= (√3 – 1)/2√2
(iii) tan 105° = tan (60° + 45°)
= tan60° + tan45°/1 – tan60° tan45°
= √3 + 1/1 + √3
(iv) cot 225° = cot (180° + 45°)
= cot 45° = 1
(2) Prove the following:
(i) Cos (π/2 – x) cos (π/2 – y) – sin (π/2 – x) sin (π/2 – y)
= – cos (x + y)
Solution:
cos (π/2 – x) cos (π/2 – y) – sin (π/2 – x)sin (π/2 – y)
= cos (π/2 – x + π/2 – y) = cos (π – (x + y)
= – cos (x + y) = RHS
(ii) tan (π/4 + θ) = 1 – tanθ/1 + tanθ
(iii) (1 + tanx/1 – tanx)2 = tan (π/4+x)/tan (π/4 – x)
(iv) sin [(n+1)A].sin[(n+2)A]+cos[(n+1)A]. cos[(n+2)A]= cos A
Solution:
(v) √2 cos (π/4 – A)
= √2 {cos π/4 cos A + sin π/4 sin A}
= √2 {1/√2 cosA + 1/√ sin A}
= cos A + sinA = RHS
(vi) cos(x – y)/cos(x + y) = cotx coty + 1/cotx coty – 1
Solution:
(vii) Cos (x + y). cos (x – y) = cos2y – sin2x
Solution:
Cos (x + y) cos (x – y)
= {cos x cos y – sinx siny} {cos x cos y + sin x sin y}
= cos2x cos2y – sin2x sin2y
= cos2y (1 – sin2x) – sin2x (1 – cos2y)
= cos2y – sin2x – sin2x cos2y + sin2x cos2y
= cos2y – sin2x = RHS
(viii) tan5A – tan3A/tan5A + tan3A = sin2A/sin8A
Solution:
(ix) tan8θ − tan5θ − tan3θ = tan8θ tan5θ tan3θ
Solution:
tan8θ – tan5θ – tan3θ
∴ 8θ = 5θ + 3θ
= tan8θ = tan (5θ + 3θ)
= tan5θ + tan 3θ/1 tan 5θ tan 3θ
= tan 8θ – tan3θ tan 5θ tan 8θ = tan5θ + tan 3θ
= tan 8θ – tan 5θ – tan 3θ = tan 8θ tan 5θ tan 3θ
(x) tan50° = tan40° + 2 tan10°
Solution:
tan 50° = tan (40° + 10°)
= (tan 40° + tan 10°)/1 – tan 40° tan 10°
=> tan 50° – tan 40° tan 10° tan 50° = tan 40° + tan 10°
=> tan 50° – tan 40° tan 10° tan (90° – 40°) = tan 40° + tan 10°
=> tan 50° – tan 40° tan 10° cot 40° = tan 40° + tan 10°
=> tan 50° = tan 40° + tan 10° + tan 10°
= tan 40° + 2tan 10°
(xi) cos27° + sin27°/cos27° – sin27° = tan 72°
Solution:
(xii) tan10° + tan35° + tan10°.tan35° = 1
Solution:
tan10° + tan35° + tan10° tan 35°
tan45° = tan (35° + 10°)
= tan 35° + tan 10°/1 – tan 35° tan 10° = 1
= tan 35° + tan 10° = 1 – tan 35° tan 10°
= tan 35° + tan 10° + tan 35° tan 10° = 1
(xiii) CotA cot4A + 1/cotA cot4A – 1 = cos3A/cos5A
(xiv)
(3) If sin A = -5/13’ π < A < 3π/2 and cosB = 3/5, 3π/2 < B < 2π
Find:
(i) Sin (A + B)
(ii) Cos (A – B)
(iii) Tan (A + B)
Solution:
Sin A = -5/1
Cos A = √1 – sin2A = √1 – 25/169 = ± √144/169 = ± 12/13
∵ π < A < 3π/2, CosA = -12/13, tanA = 5/12
CosB = 3/5
SinB = √1 – cos2B = √1 – 9/25 = ± √16/25 = ± 4/5
∵ 3π/2 < B < 2π, sin B = -4/5, tan B = – 4/3
(i) Sin (A + B) = sin A cos B + cos A sin B
= 5/13 × 3/5 – 12/13 × (-4/5)
= 3/13 + 48/13×5 = -15+48/65 = -33/65
(ii) Cos (A – B) = Cos A Cos B + sin A sin B
= – 12/13 × 3/5 + (-5/13) × (-4/5)
= -36+20/65 = -16/65
(iii) Tan (A + B)
Exercise – 3.2
(1) Find the value of:
(i) Sin 690°
(ii) Sin (495°)
(iii) cos 315°
(iv) cos (600°)
(v) Tan 225°
(vi) Tan (- 690°)
(vii) Sec 240°
(viii) Sec (- 855°)
(ix) Cosec 780°
(x) Cot (-1110°)
Solution:
(i) Sin (690°) = Sin (2×360° – 30°)
= – sin 30° = – ½
(ii) Sin (495°) = Sin (3×180° – 45°)
= Sin 45° = 1/√2
(iii) cos 315° = Cos (360° – 45°) = Cos 45° = 1/√2
(iv) cos (600°) = Cos (3×180° + 60°) = Cos 60° = – 1/2
(v) Tan 225° = Tan (180° + 45°) = tan 45° = 1
(vi) Tan (- 690°) = – Tan (690°) = – Tan (2×360° – 30°) = tan 30° = 1/√3
(vii) Sec 240° = Sec (180° + 60°) = -sec 60° = -2
(viii) Sec (- 855°) = Sec (5×180° – 45°) = – sec 45° = – √2
(ix) Cosec 780° = cosec (2×360° + 60°) = cosec 60° = 2/√3
(x) Cot (-1110°) = – cot (1110°) = -cot (3×360° + 30°) = -cot 30° = – √3
(2) Prove the following:
(i) Cos (π + x) cos (-x)/sin (π – x) cos (π/2 + x) = cot2 x
Solution:
(ii) Cos (3π/2 + x) cos (2π + x) [cot (3π/2 – x) + cot (2π + x)] = 1
Solution:
Cos (3π/2 + x) cos (2π + x) [cot (3π/2 – x) + cot (2π + x)]
= sin x cos x [tan x + cot x]
= sin x cos x [sin x/cos x + cos x/sin x]
= sin2 x + cos2 x = 1 = RHS
(iii) Sec 840°. Cot (-945°) + sin 600° tan (-690°) = 3/2
Solution:
Sec 840° – cot (-945°) + sin 600° tan (-690°)
= sec (5×180° – 60°) {- cot (5×180° + 45°)} + sin (3×180° + 60°) {- tan (4×180° – 30°)}
= sec 60° (-cot45°) + (-sin 60°) (-tan 30°)
= -2. (-1) + (- √3/2) (1/√3) = 2 – 1/2 = 3/2 = RHS
(iv) cosec (90° – x) sin (180° – x) cot (360° – x)/sec (180° + x) tan (90° + x) sin (-x) = 1
(v) sin3 (π + x) sec2 (π – x) tan (2π – x)/cos2 (π/2 + x) sin (π – x) cosec2 – x = tan3 x
(vi) cosθ + sin (270° + θ) sin (270° – θ) + cos (180° + θ) = 0
Exercise – 3.3
(1) Find values of:
(i) Sin π/8
(ii) Cos π/8
Solution:
(i) Sin π/8
Sin π/4 = sin (2× π/8) = 2 sin π/8 cos π/8
= 2 sin π/8 √1 – sin2 π/8 = 1/√2
= 4 sin2 π/8 (1 – sin2 π/8) = 1/2 (squaring both sides)
= 8 sin4 π/8 – 8 sin2 π/8 + 1 = 0
∴ sin2 π/8 = 8 ± √64 – 32/16
= 8 ± √32/16 = 2 ± √2/4 = √2 ± 1/2√2
∵ sin π/8 is positive
(ii) Cos π/8
cos π/4 = 2 cos2 π/8 – 1
= 2 cos2 π/8 = 1/√2 + 1 = cos2 π/8 = √2 + 1/2√2
(2) Find sin 2x, cos 2x, tan 2x if secx = -13/5, π/2 < x < π
Solution:
Sec x = -13/5
cos x = -5/13
sin x = √1 – cos2 x = √1 – 25/169 = √144/169 = ± 12/13 =12/13 (π/2 < x < π)
tan x = (12/13)/-5/13 = 12/5
sin 2x = 2sin x cos x = 2 (12/13) (-5/13) = -120/169
cos 2x = 2cos2 x – 1 = 50/169 – 1 = -119/169
tan 2x = (-120/169)/-119/169 = 120/119
(3) Prove the following:
(i) 1 – cos2θ/1 + cos2θ = tan2θ
Solution:
1–cosθ/1+cos2θ = 1–1+2sin2θ/1+2cos2θ = sin2θ/cos2θ = tan2θ = RHS
(ii) (sin 3x + sin x) sin x + (cos 3x-cos x) cos x = 0
Solution:
(sin 3 x + sin x) sin x + (cos 3x – cos x) cos x
= 3 (sin x – 4 sin3 x + sin x) sin x + (4cos3 x – 3cos x – cos x) cos x
= 4 (sin x – sin3 x) sin x + 4 (cos3 x – cos x) cos x
= 4 sin2x cos2x + 4cos2x (1 – sin2x – 1)
= 4 sin2x cos2x – 4sin2x cos2x = 0 = RHS
(iii) (cos x + cos y)2 + (sin x – sin y)2 = 4cos2 (x+y)/2
Solutin:
(Cosx + cosy)2 + (sinx – siny)2
= cos2x + cos2y + 2cosx cosy + sin2x + sin2y – 2sinx siny
= 1 + 1 + 2 (cosx cosy – sinx siny)
= 2 [1+ cos (x + y)] = 2 [1 + 2cos2 (x+y)/2 – 1]
= 4 cos2 x+y/2 = RHS
(iv) (cosx – cosy)2 + (sinx – siny)2 = 4sin2 (x+y)/2
Solution:
(cosx – cosy)2 + (sinx – siny)2
= cos2x + cos2y – 2 cosx cosy + sin2x + sin2y – 2sinx siny
= 1 + 1 – 2 (cosx cosy + sinx siny)
= 2 [1 – cos (x – y)]
= 2 [1 – 1 + 2 sin2 (x-y)/2]
= 4 sin2 (x-y)/2 = RHS
(v) tanx + cotx = 2 cosec2x
Solution:
tanx + cotx
= sinx/cosx + cosx/sinx
= sin2x + cos2x/sinx cosx
= 2/2sinx cosx
= 2/sin 2x
= 2cosec 2x = RHS
(vi) cosx + sinx/cosx – sinx – cosx – sinx/cosx + sinx = 2tan 2x
Solution:
(viii) 16 sinθ cosθ cos2θ cos4θ cos8θ = sin16θ
Solution:
16 sinθ cosθ cos2θ cos4θ cos8θ
= 8 (2sinθ cosθ) cos2θ cos4θ cos8θ
= 8 sin2θ cos2θ cos4θ cos8θ = 4 (2sin2θ cosθ) cos4θ cos8θ
= 4 sin 4θ cos4θ cos8θ
= 2 (2sin4θ cos4θ) cos8θ
= 2sin8θ cos8θ = sin16θ = RHS
(xii) 1/tan3A – tanA = 1/cot3A – cotA = cot2A
Solution:
1/tan3A – tanA = 1/cot3A – cotA
= 1/tan3A – tanA – tan3A tanA/tanA – tan3A
= 1+tan3A tanA/tan3A – tanA
= 1/tan (3A – A)
= cot2A = RHS
(xiii) ) cos7° cos 14° cos28° cos 56° = sin68°/16cos83°
Solution:
cos7° cos 14° cos28° cos 56°
= 1/2sin7° × 2sin7° cos7° cos14° cos28° cos56°
= 1/2sin7° sin14° cos14° cos28° cos56°
= 1/4sin7° sin28° cos28° cos56°
= 1/8sin7° sin56° cos56°
= 1/16sin7° sin 112°
= sin (180° – 68°)/16 sin (90° – 83°)
= sin68°/16cos 83° = RHS
(xiv) sin2 (-160°)/sin2 70° + sin(180° – θ/sinθ) = sec2 20°
Solution:
sin2 (-160°)/sin2 70° + sin(180° – θ/sinθ)
= sin2 (180° – 20°)/sin2 (90° – 20°) + sinθ/sinθ
= sin2 20°/cos2 20° +1
= sin2 20° + cos2 20°/cos2 20o = 1/cos2 20° = RHS
(xv) 2cos4x + 1/2cosx + 1 = (2cosx – 1) (2cos2x – 1)
(xvi) cos2 + cos2 (x + 120°) + cos2 (x – 120°) = 3/2
Solution:
Cos2 + cos2 (x + 120°) + cos2 (x – 120°)
= cos2x + (cosx cos120° – sinx sin120°)2 + (cosec cos120° + sinx sin120°)2
= cos2 x + 2 [cos2x cos2120° + sin2x sin2120°]
= cos2x + 2 [cos2 x cos2 120° + (1 – cos2x) sin2120°]
= cos2x [1 + 2 (cos2 120° – sin2 120°)] + 2 sin2 120°
= cos2x [1 + 2cos 240] +2 sin2 (180° – 60°)
= cos2x [1+2 cos (180° + 60°)] + 2 sin2 60°
= cos2 x [1 – 2 × ½] + 2 × 3/4
= cos2 x × 0 + 3/2 = 3/2 RHS
(xviii) 4cosx cos (x + π/3) + cos2 (π – 2π/3) = cos3x
Solution:
4cosx cos (x + π/3) . cos (π – 2π/3)
= 4cosx (cosx cos π/3 – sinx sin π/3) (cosx cos π/3 + sinx sin π/3)
= 4 cos x (cos x 1/2 – sin x √3/2) (cosx 1/2 + sinx √3/2)
= cosx (cosx – √3 sin x) (cos x + √3 sinx)
= cosx (cos2x – 3 sin2x)
= cos3x – 3 (1 – cos2x) cos x
= cos3x – 3 cosx + 3cos3 x
= 4 cos3x – 3cosx = cos3x = RHS
(xix) sinx tan (x/2) + 2cosx = 2/1 + tan2 (x/2)
Solution:
sinx tan (x/2) + 2cosx
= (2 sin x/2 cos x/2) ((sin x/2)/(cos x/2)) + 2cos x
= 2 sin2 x/2 + 2cos x
= 1 – cosx + 2cosx
= 1 + cosx = 2 cos2 x/2 = 2/(sec2 x/2) = 2/(1 + tan2 (x/2)) = RHS
Exercise – 3.4
(1) Express the following as a sum or difference of two trigonometric function.
(i) 2sin 4x cos 2x
Solution:
(i) 2sin 4x cos 2x
= sin (4x + 2x) + sin (4x – 2x)
= sin 6x + sin 2x
(ii) 2sin 2π/3 cos π/2
Solution:
2 sin 2π/3 cos π/2
= sin (2π/3 + π/2) + sin (2π/3 – π/2)
= sin (4π + 3π/6) + sin (4π – 3π/6)
= Sin 7π/6 + sin π/6
(iii) 2cos4θ cos2θ
Solution:
2cos 4θ cos2θ
= cos (4θ + 2θ) + cos (4θ – 2θ)
= cos 6θ + cos2θ
(iv) 2cos35° cos75°
Solution:
2cos35° cos75° = 2cos75° cos35°
= cos (75° + 35°) + cos (75° – 35°)
= cos 110° + cos 40°
(2) Prove the following:
(iv) sin18° cos39° + sin6° cos15° = sin24° cos33°
Solution:
Sin 18° cos39° + sin6° cos15°
= 1/2 (2sin 18° cos39° + 2sin 6° cos15°)
= ½ (sin (18° + 39°) – sin (39° – 18°) + sin (6° + 15°) – sin (15° – 6°))
= 1/2 (sin 57° – sin21° + sin21° – sin 9°)
= 1/2 (sin57° – sin9°)
= 1/2 (2 57°+ 9°/2 sin 57°-9°/2)
= cos 33° sin 24° = RHS
(v) cos20° cos40° cos60° cos80° = 1/16
Solution:
cos20° cos40° cos60° cos80°
= cos20° cos 40° 1/2 cos 80°
= 1/2 × 1/2 (2cos 20° cos 40°) cos 80°
= 1/4 (cos (40° + 20°) + cos (40° – 20°)) cos 80°
= 1/4 (cos 60° + cos 20°) cos 80°
= 1/8 2cos 20° cos 80° + 1/8 cos 80°
= 1/8 (cos (80° + 20°) + cos (80° – 20°)) + 1/8 cos 80°
= 1/8 (cos 100° + cos 60°) + 1/8 cos 80°
= 1/16 + 1/8 cos 100° + 1/8 cos 80°
= 1/16 + 1/8 (cos (180 – 80°) + cos 80°)
= 1/16 + 1/8 (cos 80° + cos 80°)
= 1/16 = RHS
(vi) sin20° sin40° sin60° sin80° = 3/16
Solution:
Sin20° sin40° sin60° sin80°
= sin20° sin40° √3/2 sin 80°
= √3/4 (2 sin 20 sin 40°) sin 80°
= √3/4 (cos (40° + 20°) – cos (40° – 20°)) sin 80°
= √3/4 (cos 60° – cos 20°) sin 80°
= √3/4 (1/2 – cos 20°) sin 80°
= √3/8 sin 80° – √3/8 (2 cos 20° sin 80°)
= √3/8 sin 80° – √3/8 (sin (80° + 20°) + sin (80° – 20°))
= √3/8 sin 80° – √3/8 (sin 100° + sin 60°)
= √3 sin 80° – √3/8 sin (180° – 80°) + √3/8 √3/2
= √3/8 sin 80° – √3/8 sin 80° + 3/16
= 3/16 = RHS
Exercise – 3.5
In △ABC, A + B + C = π show that
(1) cos2 A + cos2 B + cos2 C
= -1 -4 cosA cosB cosC
Solution:
cos2A + cos2B + cos2C
= 2cos 2A+2B/2 cos 2A-2B/2 + cos2C
= 2cos (A + B) cos (A – B) + 2cos2C -1
= 2cos (π – c) cos (A – B) + 2cos2C -1
= -2cos C [cos (A – B) – cos C] – 1
= -2cos C [cos (A – B) – cos (π – (A + B))] – 1
= -2cos C [cos (A – B) + cos (A + B)] – 1
= – 2 cos C [2 cos A cos B] – 1
= – 1 – 4 cos A cos B cos C = RHS
(2) sin A + sin B + sin C
= 4 cos A/2 cos B/2 cos C/2
Solution:
SinA + sinB + sinC
= 2sin A+B/2 cos A-B/2 + 2sin C/2 cos C/2
= 2sin (π/2 – C/2) cos (A-B/2) + 2sin C/2 cos C/2
= 2cos C/2 [cos (A-B/2) + sin (π/2 – A+B/2)]
= 2cos C/2 [cos (A-B/2) + cos (A+B/2)]
(3) cos A + cos B – cosC
= 4cos A/2 cos B/2 sin C/2 = 1
Solution:
cosA + cosB – cosC
= 2 cos (A+B/2) cos (A-B/2) – [1 – 2 sin2 c/2]
= 2 cos (π/2 – c/2) cos (A-B/2) -1 +2 sin2 c/2
= 2 sin c/2 [cos (A-B/2) + sin (π/2 – A+B/2)] – 1
= 2 sin c/2 [cos A-B/2 + cos A+B/2] -1
= 2 sin c/2 2cos A/2 cos B/2 -1
= 4 cos A/2 cos B/2 sin C/2 -1 = RHS
(4) sin2 A + sin2 B + sin2 C = 2 + 2cosA cosB cosC
Solution:
sin2A + sin2B + sin2C
= 1-cos2A/2 + 1-cos2B/2 + sin2C
= ½ [2 – (cos2A + cos2B)] + sin2c
= 1 – 1/2 (2 cos (A+B) cos (A-B) + sin2C
= 1 – cos (A+B) cos (A-B) + sin2c
= 1/2 [2 sin2c + 2 – 2cos (A + B) cos (A-B)]
= 1/2 [1 – cos 2c + 2 – 2cos (A+B) cos (A-B)]
= 1/2 [3 – 2cos2c + 1 – 2 cos2c + 1 – cos (π – c) cos (A – B)]
= 1/2 [4 – 2cos C (cos C + cos B))]
= 2 – cosC [cos (π – (A+B)) + cos (A-B)]
= 2 – cosC [cos (A – B) – cos (A+B)]
= 2 – cos C [2 cosA cos B]
= 2 – 2 cos A cos B cos C = RHS
(5) sin2 A/2 + sin2 B/2 sin2 C/2
= 1 – 2cos A/2 cos B/2 sin C/2
Solution:
sin2 A/2 + sin B/2 sin2 C/2
= 1 – cosA/2 + 1-cosB/2 – sin2 C/2
= 1 – 1/2 (cos A + cosB) – sin2 c/2
= 1 – 1/2 (2 cos A+B/2 cos A-B/2] – sin2 c/2
= 1 – sin2 c/2 – cos (π/2 – c/2) cos (A-B/2)
= 1 – sin2 c/2 – sin c/ cos A-B/2
= 1 – sin c/2 (sin c/2 + cos A-B/2)
= 1 – sin c/2 [sin (π/2 – A+B/2) + cos (A-B/2)]
= 1 – sin c/2 [cos A+B/2 + cos A-B/2]
= 1 – sin c/2.2 cos A/2 cos B/2
= 1 – 2cos A/2 cos B/2 sin c/2 = RHS
(6) Cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2
Solution:
Cot A/2 + cot B/2 + cot c/2
A + B + c = π => A + B = π – c
=> A/2 + B/2 = π/2 – c/2
=> Cot (A/2 + B/2) = cot (π/2 – c/2)
(7) tan2A + tan2B + tan2C = tan2A tan2B tan2C
Solution:
A + B + C = π
= 2A + 2B = 2π – 2C
= tan (2A + 2B) = tan (2π – 2C)
= tan 2A + tan 2B/1 – tan 2A tan 2B = – tan 2C
= tan 2A + tan 2B = -tan 2C + tan 2A tan 2B tan 2C
= tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C
(8) cos2 A + cos2 B – cos2 C = 1 – 2sinA sinB cosC
Solution:
cos2A + cos2B – cos2C
= cos2A +1/2 + cos2B + ½ – cos2C
= 1/2 [2 + cos 2A + cos 2B] – cos2C
= ½ [2 + 2cos 2A+2B/2 cos 2A-2B/2] – cos2C
= 1 + cos (A + B) cos (A – B) – cos2C
= 1 + cos (π – c) cos (A – B) – cos2C
= 1 – cos C cos (A-B) – cos2C
= 1 – cosC [cos (A-B) + cosC]
= 1 – cosC [cos (A – B) + cos (π – (A + B))]
= 1 – cos C [cos (A – B) – cos (A + B)]
= 1 – cos C [2sin A sinB]
= 1 – 2sin A sin B cos C
= RHS
MISCELLANEOUS EXERCISE – 3
(I) Select correct option from the given alternatives.
(1) The value of sin (n+1) Asin (n+2) A + cos (n+1) A cos (n+2) A is equal to
(A) sin A
(B) cosA
(C) − cos A
(D) sin2A
Solution:
sin (n + 1)A sin (n + 2)A + cos (n + 1)A cos (n + 2)A
= cos [(n + 2) A – (n + 1)A]
= cos A
A: (b) Cos A
(2) If tan A − tan B = x and cot B − cotA = y then cot (A − B) = …
(A) 1/y – 1/x
(B) 1/x – 1/y
(C) 1/x + 1/y
(D) xy/x-y
Solution:
(5) The value of cosA cos (60° − A) cos (60° + A) is equal to …..
(A) 1/2 cos 3A
(B) Cos 3A
(C) 1/4 cos 3A
(D) 4cos 3A
Solution:
Cos A cos (60° – A) cos (60° + A)
= 1/2 cosA [2 cos (60° – A) cos (60° + A)]
= 1/2 cos A [Cos (60° + A + 60° – A) + cos (60° + A – 60° + A)]
= ½ cosA [cos 120° + cos2A]
= 1/2 [- cos 60° cos A + (2 cos2A – 1) cos A]
= 1/2 [2 cos3A – cosA. ½ – cosA]
= 1/2 [4 cos3A – 3cosA] = 1/4 cos 3A
A: (C) 1/ cos 3A
(6) The value of sin π/14 sin 3π/14 sin 5π/14 sin 7π/14 sin 9π/14 sin 11π/14 sin 13π/14 is ……
(A) 1/16
(B) 1/64
(C) 1/128
(D) 1/256
Solution:
Sin π/14 sin 3π/14 sin 5π/14 sin 7π/14 sin 9π/14 sin 11π/14 sin 13π/14
= sin π/14 sin 3π/14 sin 5π/14 sin π/2 sin (π – 5π/14) sin (π – 3π/14) sin (π – π/14)
= sin π/14 sin 3π/14 sin 5π/14 sin 5π/14 sin 3π/14 sin π/14
= (sin π/14 sin 3π/14 sin 5π/14)2
= [sin (π/2 – π/7) sin (π/2 – 2π/7) sin (π/2 – 3π/7)]2
= (cos π/7 cos 2π/7 cos 3π/7)2
(7) If α + β + ϰ = π then the value of sin2 α + sin2 β-sin2 ϰ is equal to…..
(A) 2sin α
(B) 2sin α cos β sinϰ
(C) 2sin α sinβcosϰ
(D) 2sin α sinβsinϰ
Solution:
Sin2 α + sin2 β – sin2 ϰ
= 1 – cos2 α/2 + 1 – cos2 β/2 = -sin2 ϰ
= 1/2 [2 – (cos2 α + cos2β)] – sin2 ϰ
= ½ [2 – 2 cos (α + β) – cos (α – β)] – sin2 ϰ
= 1 – cos (π – ϰ) cos (α – β) – sin2 ϰ
= cos2 ϰ + cos ϰ cos (α – β)
= cos ϰ (cos ϰ + cos (α – β))
= cos ϰ [cos (π – (α – β)) + cos (α – β)]
= cos ϰ [cos (α – β) – cos (α – β)]
= cos ϰ [2 sin α sin β]
= 2 sin α sin β cos ϰ
A: (C) 2 sin α sin β cos ϰ
(8) Let 0 < A, B < π/2 satisfying the equation 3sin2 A + 2sin2 B = 1 and 3sin2 A – 2sin2B = 0 then A+2B is equal to….
(A) π
(B) π/2
(C) π/4
(D) 2π
Solution:
3sin 2A – 2sin 2B = 0
= sin 2B = 3/2 sin 2A —– (i)
3 sin2A + 2sin2B = 1
= 3sin2A = 1 – 2sin2B
= 3sin2A = cos 2B
cos (A + 2B) = cos A cos 2B – sinA sin2B
= cos A (3sin2A) – sinA. (3/2 sin 2A)
= 3 cos A sin2A – 3sin2A cos A
= 0
∴ A + 2B = π/2
A: (B) π/2
(9) In∆ABC if cot A cot B cotC > 0 then the triangle is….
(A) Acute angled
(B) Right angled
(C) Obtuse angled
(D) Isosceles right angled
Solution:
∵ cot A cot B cot C > 0
cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < π/2, 0 < B < π/2, 0 < C < π/2
∴ A, B, C are acute angles
A: (A) Acute angled
(10) The numerical value of tan20° tan80° cot50° is equal to…..
Solution:
(A) √3
(B) 1/√3
(C) 2√3
(D) 1/2√3
Solution:
tan20° tan80° cot50°
= tan20° tan80° cot (90° – 40°)
= tan20° tan80° tan40°
= tan20° tan [60° + 20°] tan [60° – 20°]
= tan20° [3 – tan220°/1 – 3tan2 20°]
= 3tan20° – tan320°/1 – 3 tan220° = tan (3×20°)
= tan 60° = √3
A: (A) √3
(II) Prove the following.
(1) tan20° tan80° cot50° = √3
Solution:
Same as previous question.
(2) If sin α sinβ − cos α cos β + 1= 0 then prove cot α tan β = −1
Solution:
Sin α sin β – cos α cos β + 1 = 0
= cos (α + β) = 1
= α + β = 0 = α = – β
∴ Cot α tan β = cot α tan (-α)
= cot α tan α
= -1
(3) cos 2π/15 cos 4π/15 cos 8π/15 cos 16π/15 = 1/16
Solution:
(4) (1 + cos π/8) (1 + cos 3π/8) (1 + cos 5π/8) (1 + cos 7π/8) = 1/8
Solution:
(1 + cos π/8) (1 + cos 3π/8) (1 + cos 5π/8) (1 + cos 7π/8)
= (1 + cos π/8) (1 + cos 3π/8) (1 + cos (π – 3π/8)) (1 + cos (π – π/8))
= (1 + cos π/8) (1 + cos 3π/8) (1 – cos 3π/8) (1 – cos π/8)
= (1 – cos2 π/8) (1 – cos2 3π/8)
= sin2 π/8 sin 3π/8 = 1/4 (2 sin π/8 sin 3π/8)2
= 1/4 [cos (π/8 – 3π/8) – cos (π/8 + 3π/8)]2
= 1/4 [cos (- π/4) – cos (π/2)]2
= 1/4 [1/√2 – 0]2
= 1/8 = RHS
(5) cos12°+ cos 84° + cos 156° + cos132° = -1/2
Solution:
cos 12° + cos 84° + cos 156° + cos 132°
= cos 12° + cos 84° + cos (180° – 24°) + cos (180° – 48°)
= cos 12° + cos 84° – cos 24° – cos 48°
= (cos 12° – cos 48°) – (cos 24° – cos 84°)
= 2 sin 30° sin 18° – 2 sin 54° sin 30°
= 2. 1/2 [sin 18° – sin 54°]
= sin 18° – sin3 × 18°
= sin 18° – 3sin 18° + 4sin318°
= 4sin3 18° – 2sin18°
= 2sin 18° [2sin218° -1]
= 2sin 18° cos 36°
= -1/cos18° (2 sin 18° cos 18° cos 36°)
= – 1/cos18° (sin36° cos36°)
= – 1/2cos18° sin72°
= – 1/2cos18° sin (90° – 18°)
= – 1/2cos18° cos18°
= – 1/2 = RHS
(6) cos (π/4 + x) + cos (π/4 – x) = √2 cos x
Solution:
= 2 cos π/4 cos x
= 2/√2 cos x
= √2 cos x = RHS
(7) Sin5x – 2sin3x + sinx/cos5x – cosx = tanx
Solution:
(8) Sin2 6x − sin2 4x = sin2x sin10x
Solution:
Sin2 6x − sin2 4x
= (sin6x – sin4x) (sin6x + sin4x)
= 2sinx cos 5x. 2sin 5x cos x
= (2sinx cosx) (2 sin 5x cos 5x)
= sin 2x sin 10x = RHS
(9) Cos2 2x − cos2 6x = sin4x sin8x
Solution:
Cos2 2x − cos2 6x
= (Cos 2x – cos 6x) (cos 2x + cos 6x)
= 2 sin 4x (sin 2x). 2 cos 4x cos (-2x)
= (2sin 4x cos 4x) (2 sin 2cos 2x)
= sin 4x sin 8x = RHS
(10) Cot4x ( sin5x + sin3x) = cot x (sin5 x − sin3x)
Solution:
Cot 4x (sin 5x + sin 3x)
= cos 4x/sin 4x (2 sin 4x cos x)
= 2 cos x cos4x = 2 cosx/sinx sinx cos 4x
= cot x. 2sin x cos 4x
= cot x [sin 5x – sin 3x] = RHS
(11) Cos9x – cos5x/sin17x – sin3x = sin2x/cos10x
Solution:
Cos9x – cos5x/sin17x – sin3x = -2sin7x sin2x/2sin7x cos10x = – sin2x/cos10x = RHS
(12) If sin 2A = λsin 2 B then prove that
tan (A+B)/tan (A-B) = λ+1/λ-1
Solution:
Sin 2A = λ sin 2B.
=> sin 2A/sin 2B = λ => sin2A + sin 2B/sin 2A – sin 2B = λ+1/λ-1
= PHOTO
= tan (A+B) cot (A-B) = λ+1/λ-1
= Tan (A+B)/Tan (A-B) = λ+1/λ-1
(13) 2cos2A+1/2cos2A-1 = tan(60° + A) tan (60° – A)
(15) 3tan6 10° – 27 tan4 10° + 33tan2 10° = 1
Solution:
3tan6 10° – 27 tan4 10° + 33tan2 10°
tan 30°= 1/√3 = tan 3(10°) = 1/√3
= 3tan10° – tan310°/1-3tan210° = 1/√3
= (3tan10° – tan310°)2/(1 – 3tan210°)2 = 1/3
= 3 (9tan210° + tan610° – 6tan410°) = 1 + 9 tan410° – 6tan210°
= 27tan210° + 3tan610° – 18tan410° = 1 + 9tan410° – 6tan210°
= 3tan610° – 27tan410° + 33tan210° = 1
(16) cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
cosec 48° + cosec 96° + cosec 192° + cosec 384°
Cosec 48° = cosec (90° – 42) = sec 42°
Cosec 192° = cosec (270° – 78°) = sec 78°
Cosec 384° = cosec (360° + 24°) = cosec 24°
∴ Cosec 48° + cosec 96° + cosec 192° + cosec 384°
= sec 42° – sec 78° + cosec 96° + cosec 24°
= 1/cos 42° – 1/cos 78° + 1/sin 96° + 1/sin 24°
(17) 3 (sinx – cosx)4 + 6 (sinx + cosx)2 + 4 (sin6x + cos6x) = 13
Solution:
3 (sinx – cosx)4 + 6 (sinx + cosx)2 + 4 (sin6x + cos6x)
(sinx – cosx)4 = (sinx – cosx)2)2
= (sin2x + cos2x – 2sinx cosx)2
= (1 – 2sinx cosx)2 = 1 + 4 sin2x cos2x – 4sinx cosx
(sinx + cosx)2 = sin2 + cos2x + 2sinx cosx = 1 + 2 sinx cosx
∴ 3 (sinx – cosx)4 + 6 (sinx + cosx)2 + 4 (sin6x + cos6x)
= 3 + 12 sin2x cos2x – 12sinx cosx + 6 + 12 sinx cosx + 4 (sin6x + cos6x)
= 9 + 12 sin2 cos2x + 4 (sin6x + cos6x)
cos6x + sin6x = (sin2x)3 + (cos3x)3
= (sin2x + cos2x) (sin4x + cos4x – sin2x cos2x)
= (sin2x + cos2x)2 – 2sin2x cos2x – sin2x cos2x
= 1 – 3 sin2x cos2x
∴ 3 (sinx – cosx)4 + 6 (sinx + cosx)2 + 4 (sin6x + cos6x)
= 9 + 12sin2x cos2x + 4 – 12 sin2x cos2x = 13 = RHS
(18) Tan A + 2 tan2A + 4tan4A + 8cot 8A = cotA
Solution:
Tan A + 2 tan2A + 4tan4A + 8cot 8A
= TanA + 2tan2A + 4tan 4A + 8 (1-tan24A/2tan4A)
= tanA + 2tan 2A + 4 [tan2 4A + 1 – tan2 4A/tan 4A]
= tan A + 2tan 2A + 4/Tan 4A
= tan A + 2tan 2A + 4(1-tan22A/2tan 2A)
= tanA + 2 [tan22 + 1 – tan22A/tan 2A]
= tan A + 2/tanA
= tanA + 2 (1 – tan2A)/2tanA
= tan2A + 1 – tan2A/tanA
= 1/TanA = cotA = RHS
(19) If A + B + C = 3π/2 then cos2A + cos2B + cos2C = 1 – 4sinAsinBsinC
Solution:
cos2A + cos2B + cos2C
= 2cos (A + B) cos (A – B) + cos 2C
= 2cos (3π/2 – C) cos (A – B) + cos 2C
= 2 (-sinC) cos (A – B) + 1-2sin2 C
= 1 – sin C {cos (A – B) + sin (3π/2 – (A + B))}
= 1 – 2 sin c {cos (A – B) – cos (A + B)}
= 1 – 2 sin C {2 sin A SinB}
= 1 – 4 SinA SinB sinC = RHS
(20) In any triangle ABC, sin A − cosB = cosC then ∠B = π/2
Solution:
SinA – cosB = cosC
= sinA = cosB + cosC = 2cos B+C/2 cos B-C/2
= 2cos (π-A/2) cos B-C/2
= 2sin A/2 cos (B-C/2)
= 2sin A/2 cos A/2 = 2sin A/2 cos (B-C/2)
= cos A/2 = cos (B-C/2) = A/2 = B-C/2 = A = B-C
∵ A + B + C = π = B + C + B – C = π = 2B = π = B = π/2
(21) tan3x/1 + tan2x + cot3x/1 + cot2x = secxcosecx – 2sinx cosx
Solution:
tan3x/1 + tan2x + cot3x/1 + cot2x
= tan3x/sec2x + cot3x/cosec2x = sin3x/cos3x × cos2x + cos3x/sin3x × sin2x
= sin3x/cosx + cos3x/sinx = sin4x + cos4x/sinx cosx
(22) sin20° sin40° sin80° = √3/8
Solution:
= 1/2 [cos (40° – 20°) – cos (40° + 20°)] sin80°
= 1/2 [cos 20° – cos 60°] sin 80° = 1/2 cos 20° sin 80° – 1/4 sin 80°
= 1/4 (sin (80° + 20°) + sin (80° – 20°)] – 1/4 sin 80°
= 1/4 sin 100° + ¼ sin 60° – 1/4 sin 80°
= 1/4 sin (180° – 80°) + √3/8 – 1/4 sin 80° = 1/4 sin 80° – 1/4 sin 80° + √3/8
= √3/8 = RHS
(23) sin18° = √5-1/4
Solution:
90° = 5×18° = 5θ (Let)
∴ 90° = 3θ + 2θ
= sin 2θ = sin (90° = 3θ) = cos 3θ
= 2sinθ cosθ = 4 cos3θ – 3cosθ
= 4cos3θ – 3cosθ – 2sinθ cosθ = 0
= cosθ [4cos2θ – 2sinθ – 3] = 0
= 4 – 4 sin2θ – 2sinθ – 3 = 0
= 4sin2θ + 2sinθ – 1 = 0
(24) cos36° = √5+1/4
Solution:
Cos36° – cos 2×18°
= 1 – 2sin218° = 1 – 2 (√5-1/4)2
= 1 – 5+1-2√5/8
= 8-6+2√5/8 = 2+2√5/8 = 1+√5/4
(29) sin47° + sin 61° − sin11° − sin25° = cos7°
Solution:
sin47° + sin 61° − sin11° − sin25° = cos7°
= (sin47° – sin25°) + (sin61° – sin11°)
= 2cos 36° sin11° + 2cos36° sin25°
= 2cos36° (sin11° + sin25°)
= 2cos 36° [2sin 18° cos7°]
= 2 × √5+1/4 × 2 × √5-1/4 cos7°
= 5-1/4 cos7° = cos7° = RHS
(30) √3 cosec 20° – sec20° = 4
Solution:
(31) In △ABC, ∠C = 2π/3 then prove that cos2A + cos2B – cosAcosB = ¾
Solution:
cos2A + cos2B – cosA cosB
= cos2A + 1 – sin2B – cosA cosB
= 1 + (cos2A – sin2B) – cosA cosB = 1 + cos (A+B) cos (A-B) – cosA cosB
= 1 + cos (π – c) cos (A-B) – cos A cos B
= 1 + cos (π – 2π/3) cos (A – B) – cosA cos B
= 1 + ½ cos (A-B) – cosA cosB
= 1 + 1/2 (cosA cosB + sinA sinB) – cosA cosB
= 1 + 1/2 (sinA sinB – cosA coB)
= 1 – 1/2 cos (A+B)
= 1 – 1/2 cos (π – c) = 1 – 1/2 cos π/3 = 1 – ¼ = 3/4 = RHS