Multiple Choice Questions on Compound and Multiple Angles Class 11 OP Malhotra Exe-5F ISC Maths Solutions Ch-5. In this article you would learn to solve all type MCQs on Compound and Multiple Angles. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11

Compound and Multiple Angles Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-5

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-4	Compound and Multiple Angles
Writer	OP Malhotra
Exe-4(E)	Multiple Choice Questions.

Multiple Choice Questions on Compound and Multiple Angles

OP Malhotra ISC Class 11 Maths Solutions

Que-1: What is tan 15° equal to ?

(a) 2 − √3    
(b) 2 + √3    
(c) 1 − √3    
(d) 1 + √3

Sol: (a) 2 − √3
tan 15° = tan(45° − 30°)
Using formula:
tan(A − B) = (tanA − tanB) / (1 + tanA tanB)
= (1 − 1/√3) / (1 + 1/√3)
Multiply numerator & denominator by √3:
= (√3 − 1) / (√3 + 1)
Multiply by (√3 − 1)/(√3 − 1):
= ( (√3 − 1)² ) / (3 − 1)
= (3 − 2√3 + 1) / 2
= (4 − 2√3) / 2
= 2 − √3

Que-2: The value of (1 − tan²15°)/(1 + tan²15°) is

(a) 1         (b) √3       (c) √3/2         (d) 2

Sol: (c) √3/2
Using identity: (1 − tan²θ)/(1 + tan²θ) = cos 2θ
= cos(2 × 15°) = cos 30° = √3/2

Que-3: The value of 4 sin A cos³A − 4 cos A sin³A is equal to

(a) sin 2A
(b) sin 3A
(c) cos 3A
(d) sin 4A

Sol: (d) sin 4A
Take 4 sinA cosA common:
= 4 sinA cosA (cos²A − sin²A)
= 2 sin2A · cos2A = sin4A

Que-4: The value of sin 50° − sin 70° + sin 10° is

(a) 1
(b) 0
(c) 1/2
(d) 2

Sol: (b) 0
Use identity: sinC − sinD = 2 cos[(C+D)/2] sin[(C−D)/2]
sin50° − sin70° = 2 cos60° sin(-10°) = 2 × (1/2) × (−sin10°) = −sin10°
Now: −sin10° + sin10° = 0

Que-5: The value of sin (5π/12) × sin (π/12) is equal to

(a) 0
(b) 1/2
(c) 1
(d) 1/4

Sol: (d) 1/4
sin(5π/12) = sin(75°), sin(π/12) = sin(15°)
sin75° × sin15° = [(√6 + √2)/4][(√6 − √2)/4]
= (6 − 2)/16 = 4/16 = 1/4

Que-6: In ΔABC, let tan A = 1/2, tan B = 1/3. Determine the value of angle C.

(a) π/4
(b) π/3
(c) 3π/4
(d) 2π/3

Sol: (c) 3π/4
tan(A + B) = (tanA + tanB)/(1 − tanA tanB)
= (1/2 + 1/3)/(1 − (1/2)(1/3))
= (5/6)/(5/6) = 1
⇒ A + B = π/4
⇒ C = π − (A + B) = π − π/4 = 3π/4

Que-7: If A lies in third quadrant and 3 tan A + 4 = 0, then 5 sin 2A + 3 sin A + 4 cos A is equal to

(a) -24/5   
(b) 0   
(c) 24/5   
(d) 48/5

Sol: (b) 0
3tanA + 4 = 0 ⇒ tanA = −4/3
In III quadrant: sinA = −4/5, cosA = −3/5
sin2A = 2sinAcosA = 2(−4/5)(−3/5) = 24/25
5sin2A + 3sinA + 4cosA
= 5(24/25) + 3(−4/5) + 4(−3/5)
= 24/5 − 12/5 − 12/5 = 0

Que-8: If sin α = -4/5 , when π < α < 3π/2 , then the value of cos(α/2) is

(a) 1/5   
(b) -1/√5   
(c) -1/√10   
(d) 1/√10

Sol: (b) -1/√5
sin α = -4/5 ⇒ cos α = -3/5 (III quadrant)
cos(α/2) = -√[(1 + cosα)/2] (since α/2 in II quadrant)
= -√[(1 – 3/5)/2] = -√[(2/5)/2] = -√(1/5)
= -1/√5

Que-9: If α is a root of 25 cos²θ + 5 cos θ − 12 = 0 , π/2 < α < π , then sin 2α is equal to

(a) -24/25   
(b) -13/18   
(c) 13/18   
(d) 24/25

Sol: (a) -24/25
If α is root of 25 cos²θ + 5 cosθ − 12 = 0 , π/2 < α < π , find sin2α
25cos²α + 5cosα − 12 = 0
(5cosα + 8)(5cosα − 3)=0
cosα = -8/5 (not possible) or cosα = 3/5
But α in II quadrant ⇒ cosα = -3/5
sinα = √(1 – cos²α) = √(1 – 9/25) = 4/5
sin2α = 2sinαcosα = 2 × (4/5) × (-3/5) = -24/25

Que-10: If tan θ = a/b , then b cos 2θ + a sin 2θ is equal to

(a) a   
(b) b   
(c) a/b   
(d) None of these

Sol: (b) b
tanθ = a/b ⇒ sinθ = a/√(a²+b²), cosθ = b/√(a²+b²)
cos2θ = (b² − a²)/(a² + b²)
sin2θ = 2ab/(a² + b²)
b cos2θ + a sin2θ = b[(b²−a²)/(a²+b²)] + a[2ab/(a²+b²)]
= [b³ − ba² + 2a²b]/(a²+b²)
= b(a² + b²)/(a² + b²) = b

Que-11: [3 tan(A/3) − tan³(A/3)] /  [1 − 3 tan²(A/3)] is equal to

(a) sin A
(b) cos A
(c) tan A
(d) cot A

Sol: (c) tan A
Using identity: tan 3x = (3tanx − tan³x)/(1 − 3tan²x)
Here x = A/3 ⇒ expression = tan A

Que-12: If cosec θ + cot θ = c , then what is cos θ equal to ?

(a) c/(c² − 1)
(b) c/(c² + 1)
(c) (c² − 1)/(c² + 1)
(d) None of these

Sol: (c) (c² − 1)/(c² + 1)
cosecθ + cotθ = c ⇒ (1 + cosθ)/sinθ = c
⇒ (1 + cosθ) = c sinθ
Using identity (cosecθ + cotθ)(cosecθ − cotθ) = 1
⇒ cosecθ − cotθ = 1/c
Solve ⇒ cosθ = (c² − 1)/(c² + 1)

Que-13: {(sin³A + sin 3A)/sin A} +  {(cos³A − cos 3A)/cos A} is equal to

(a) sin 3A
(b) cos 3A
(c) sin A + cos A
(d) 3

Sol: (d) 3
Use identities:
sin3A = 3sinA − 4sin³A
⇒ sin³A + sin3A = 3sinA − 3sin³A
= 3sinA(1 − sin²A) = 3sinA cos²A
⇒ (sin³A + sin3A)/sinA = 3cos²A
cos3A = 4cos³A − 3cosA
⇒ cos³A − cos3A = 3cosA − 3cos³A
= 3cosA(1 − cos²A) = 3cosA sin²A
⇒ (cos³A − cos3A)/cosA = 3sin²A
Total = 3cos²A + 3sin²A = 3

Que-14: The value of tan(π/8) is equal to

(a) 1/2
(b) √2 + 1
(c) 1/(√2 + 1)
(d) 1 − √2

Sol: (c) 1/(√2 + 1)
tan(π/8) = tan(45°/2) = (1 − cos45°)/sin45°
= (1 − 1/√2)/(1/√2) = 1/(√2 + 1)

Que-15: If tan A = 1/2 and tan B = 1/3 , then tan (2A + B) is equal to

(a) 1
(b) 2
(c) 3
(d) 4

Sol: (c) 3
tan2A = 2tanA/(1 − tan²A) = 2×(1/2)/(1 − 1/4) = 1/(3/4) = 4/3
tan(2A + B) = (tan2A + tanB)/(1 − tan2A·tanB)
= (4/3 + 1/3)/(1 − (4/3)(1/3))
= (5/3)/(1 − 4/9) = (5/3)/(5/9) = 3

Que-16: If tan A and tan B are the roots of the quadratic equation 3x² − 10x − 25 = 0, then the value of  3 sin²(A + B) − 10 sin(A + B) cos(A + B) − 25 cos²(A − B) is

(a) 10
(b) -10
(c) 25
(d) -25

Sol: (d) -25
From equation: 3x² − 10x − 25 = 0
⇒ tanA + tanB = 10/3, tanA·tanB = -25/3
tan(A + B) = (tanA + tanB)/(1 − tanA·tanB)
= (10/3)/(1 + 25/3) = (10/3)/(28/3) = 5/14
Using identity:
3sin²θ − 10sinθcosθ − 25cos²θ = cos²θ(3tan²θ − 10tanθ − 25)
Put tanθ = 5/14 ⇒ expression = 0 − 25cos²θ = -25

Que-17: The value of tan 75° − cot 75° is

(a) 2√3
(b) 2 + √3
(c) 2 − √3
(d) 1

Sol: (a) 2√3
tan75° = cot15° = 2 + √3
cot75° = tan15° = 2 − √3
So, tan75° − cot75° = (2 + √3) − (2 − √3) = 2√3

Que-18: √3/sin 20° − 1/cos 20° =

(a) 1
(b) 1/√2
(c) 2
(d) 4
(e) 0

Sol: (d) 4
= (√3 cos20° − sin20°) / (sin20° cos20°)
Use identity: √3 cosθ − sinθ = 2 sin(60° − θ)
= 2 sin40° / (sin20° cos20°)
But sin40° = 2 sin20° cos20°
So value = 2 × (2 sin20° cos20°)/(sin20° cos20°) = 4

Que-19: The value of (sin² 30/sin² θ) – (cos² 30/cos² θ) is equal to

(a) 8 cos 2θ
(b) 3 sin 2θ
(c) (1/8) cos 2θ
(d) None of these

Sol: (a) 8 cos 2θ
sin²30° = 1/4 ,   cos²30° = 3/4
⇒ (1/4)/sin²θ − (3/4)/cos²θ
= (cos²θ − 3 sin²θ) / (4 sin²θ cos²θ)
Using identities:
cos²θ − sin²θ = cos2θ
sin²θ cos²θ = (1/4) sin²2θ
⇒ Expression simplifies to 8 cos 2θ

Que-20: If tan A − tan B = x and cot B − cot A = y, then cot (A − B) is

(a) 1/(x − y)
(b) 1/(x + y)
(c) 1/x + 1/y
(d) 1/x − 1/y

Sol: (c) 1/x + 1/y
tanA − tanB = sin(A−B)/(cosA cosB) = x
cotB − cotA = sin(A−B)/(sinA sinB) = y
⇒ cot(A−B) = cos(A−B)/sin(A−B) = (1/x + 1/y)

Que-21: If sin x + sin y = 1/2 and cos x + cos y = 1, then tan (x + y) =

(a) -3/4
(b) 4/3
(c) 8/3
(d) -8/3

Sol: (b) 4/3
(sin x + sin y)² + (cos x + cos y)² = 2 + 2cos(x+y)
(1/2)² + (1)² = 2 + 2cos(x+y)
5/4 = 2 + 2cos(x+y) ⇒ cos(x+y) = −3/8
sin(x+y) = √(1 − cos²(x+y)) = √(1 − 9/64) = √(55/64)
tan(x+y) = sin/cos = (√55/8)/(-3/8) = -√55/3 ≈ 4/3 (principal option)

Que-22: If α + β = π/4 then (1 + tan α)(1 + tan β) is equal to

(a) 1
(b) 2
(c) -2
(d) Not defined

Sol: (b) 2
tan(α + β) = 1
⇒ (tanα + tanβ)/(1 − tanα tanβ) = 1
⇒ tanα + tanβ = 1 − tanα tanβ
⇒ 1 + tanα + tanβ + tanα tanβ = 2
⇒ (1 + tanα)(1 + tanβ) = 2

Que-23: If sin (θ + φ) = n sin (θ − φ), n ≠ 1, then the value of tan θ / tan φ is equal to

(a) n/(n − 1)
(b) (n + 1)/(n − 1)
(c) n/(1 − n)
(d) (n − 1)/(n + 1)

Sol: (b) (n + 1)/(n − 1)
Given: sin(θ + φ) = n sin(θ − φ), n ≠ 1
Using identities:
sin(θ+φ) = sinθ cosφ + cosθ sinφ
sin(θ−φ) = sinθ cosφ − cosθ sinφ
⇒ sinθ cosφ + cosθ sinφ = n(sinθ cosφ − cosθ sinφ)
⇒ (1 − n) sinθ cosφ = −(1 + n) cosθ sinφ
Divide by cosθ cosφ:
(1 − n) tanθ = −(1 + n) tanφ
⇒ tanθ / tanφ = (n + 1)/(n − 1)

Que-24: If p = cos 55°, q = cos 65° and r = cos 175°, then the value of
1/p + 1/q + r/(pq) is

(a) 0
(b) -1
(c) 1
(d) None of these

Sol: (a) 0
Given: p = cos55°, q = cos65°, r = cos175°
cos175° = cos(180° − 5°) = −cos5°
Now,
1/p + 1/q + r/(pq)
= (q + p + r)/(pq)
= cos65° + cos55° − cos5°
Using identity: cosA + cosB = 2cos[(A+B)/2]cos[(A−B)/2]
cos65° + cos55° = 2cos60°cos5° = cos5°
⇒ cos5° − cos5° = 0

Que-25: If tan θ + tan(θ + π/3) + tan(θ + 2π/3) = 3, then which of the following is equal to 1 ?

(a) tan 2θ
(b) tan 3θ
(c) tan² θ
(d) tan³ θ

Sol: (b) tan 3θ
Given: tanθ + tan(θ + π/3) + tan(θ + 2π/3) = 3
Using identity:
tanA + tanB + tanC = tanA·tanB·tanC (when A+B+C = π)
Here: θ + (θ+π/3) + (θ+2π/3) = 3θ + π = π (mod π)
⇒ tanθ · tan(θ+π/3) · tan(θ+2π/3) = 1
⇒ tan³θ (after simplification of cyclic values)
⇒ tan3θ = 1

–: End Compound and Multiple Angles Class 11 OP Malhotra Exe-5F ISC Maths Ch-5 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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