Maharashtra Board Class 11 Maths Chapter 3 Trigonometry 2 Solution

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

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

Updated: November 15, 2021 — 7:49 pm

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