JEE Advanced Maths Trigonometric Equations Previous Year Questions with Solutions

Students can find JEE Advanced previous year questions and solutions on trigonometric equations on this page. Students can expect questions from the trigonometric equations for the JEE Advanced exam. The important topics in trigonometric functions are solutions of trigonometric equations, trigonometric identities, greatest and least value of trigonometric expressions, domain and range of trigonometric functions, etc. Practising these solutions will definitely help students to get a better idea of the difficulty levels of the JEE Advanced exam.

Download Trigonometric Equations Previous Year Solved Questions PDF

Question 1: The positive integer value of n>3 satisfying the equation 1/sin(π/n) = 1/sin(2π/n) + 1/sin(3π/n)

(a) 8

(b) 6

(c) 5

(d) 7

Solution:

Given, 1/sin (π/n) = 1/sin (2π/n) + 1/sin (3π/n)

(1/sin (π/n) – 1/sin (3π/n)) = 1/sin (2π/n)

Using the formula sin x – sin y = 2 cos(x + y) sin(x – y),

Using the formula 2 sin A cos A = sin 2A,

sin (4π/n)/ sin (3π/n) = 1

sin (4π/n) = sin (3π/n)

sin(π – (4π/n)) = sin 3π/n

π – (4π/n) = 3π/n

π = (3π/n) + (4π/n)

π = 7π/n

So, n = 7

Hence option d is the answer.

Question 2: If A + B + C = 180⁰ then the value of tan A + tan B + tan C is

(a) ≥3√3

(b) ≥2√3

(c) > 3√3

(d) > 2√3

Solution:

Given A + B + C = 180⁰

So A + B = 180 – C

tan (A + B) = tan (180 – C)

(tan A + tan B)/(1 – tan A tan B) = -tan C

=> (tan A + tan B) = -tan C (1 – tan A tan B)

=> tan A + tan B + tan C = tan A tan B tan C …(i)

Use A.M ≥ G.M

=> (tan A + tan B + tan C)/3 ≥ (tan A tan B tan C)^1/3

=> tan A tan B tan C ≥ 3 (tan A tan B tan C)^1/3 (using (i))

Cubing both sides

tan²A tan²B tan²C ≥ 27

tan A tan B tan C ≥ 3√3

Hence option a is the answer.

Question 3: The number of values of θ in the interval (-π/2, π/2) such that θ ≠ nπ/5 for n = 0, ±1, ±2 and tan θ = cot 5θ as well as sin 2θ = cos 4θ

(a) 3

(b) 4

(c) 7

(d) 5

Solution:

Given tan θ = cot 5θ

= tan (π/2 – 5θ)

=> θ = nπ + π/2 – 5θ

=> 6θ = nπ + π/2

=> θ = nπ/6 + π/12 …(i)

sin 2θ = cos 4θ

=> 2 sin² 2θ + sin 2θ – 1 = 0

=> 2 sin² 2θ + 2sin 2θ – sin 2θ – 1 = 0

=> sin 2θ = -1 or sin 2θ = ½

=> 2θ = π/6 , 5π/6

=> θ = π/12 , 5π/12

So θ = -π/4, π/12 , 5π/12

θ takes 3 values.

Hence option a is the answer.

Question 4: For x ∈(0, π), the equation sin x + 2 sin 2x – sin 3x = 3 has

(a) infinitely many solutions

(b) three solutions

(c) one solution

(d) no solution

Solution:

Given that sin x + 2 sin 2x – sin 3x = 3 …(i)

Use sin 2x = 2 sin x cos x

And sin 3x = 3 sin x – 4 sin³ x

Equation (i) becomes sin x + 4 sin x cos x – 3 sin x + 4 sin³ x = 3

sin x(1 + 4 cos x – 3 + 4 sin²x) = 3

sin x(-2 + 4 cos x + 4(1-cos²x)) = 3

(-2 + 4 cos x + 4(1-cos²x)) = 3/sin x

(-2 + 4 cos x + 4(1-cos²x)) = 3 cosec x

-2 + 4 cos x + 4 – 4cos²x = 3 cosec x

(2 – (4 cos²x – 4 cos x +1) + 1) = 3 cosec x

3 – (2 cos x -1)² = 3 cosec x

When x = π/2, RHS = 3

LHS = 3, at x = π/3.

LHS and RHS are not equal at the same value of x. Hence, no solution.

Hence option d is the answer.

Question 5: If 5(tan² x – cos² x) = 2cos 2x + 9, then the value of cos 4x is :

(a) 1/3

(b) 2/9

(c) -7/9

(d) -3/5

Solution:

Given that 5(tan² x – cos² x) = 2cos 2x + 9

(Use tan²x = sec²x – 1 and cos 2x = 2 cos²x – 1)

5(sec²x – 1 – cos² x) = 2(2 cos²x – 1) + 9

5 sec² x – 5 = 9 cos² x + 7

Let cos² x = t

(5 / t) = 9t + 12

9t² + 12t – 5 = 0

t = 1/3 or – 5/3

t = 1/3 as t ≠ – 5/3

cos² x = 1/3, cos 2x = 2cos² x – 1 = -1/3

cos 4x = 2cos² 2x – 1

= (2/9) – 1

= -7/ 9

Hence option c is the answer.

Question 6: If √2 sin α/√(1 + cos 2α) = 1/7 and √((1 – cos 2β)/2) = 1/√10, α, β ∈ (0, π/2), then tan (α + 2β) is equal to

(a) 1

(b) -1

(c) 0

(d) 1/2

Solution:

We know cos 2x = 2 cos²x – 1

Also 1+cos 2x = 2 cos²x

Given that √2 sin α/√(1 + cos 2α) = 1/7

=> √2 sin α/√(2cos² α) = 1/7

=> tan α = 1/7

Given √((1 – cos 2β)/2) = 1/√10

Use cos 2x = cos²x – sin²x in above equation

=> √((1 – cos²β + sin²β)/2) = 1/√10

=> √((sin²β + sin²β)/2) = 1/√10

=> √((2 sin²β)/2) = 1/√10

=> sin β = 1/√10

=> tan β = 1/3

We know tan 2β = (2 tan β)/(1- tan²β)

= 2(1/3)/(1 – 1/9)

= 3/4

We know tan (a+b) = (tan a + tan b)/(1 – tan a tan b)

So tan(α + 2β) = (tan α + tan 2β)/( 1 – tan α tan 2β)

= (1/7 + ¾)/(1 – (1/7) × (3/4))

= 25/25

= 1

Hence option a is the answer.

Question 7: Let the function f: (0, π) → R be defined by f(θ) = (sin θ + cos θ)² + (sin θ – cos θ)⁴ . Suppose, the function f has a local minimum at θ precisely when θ ∈ {λ₁ π, … , λ_r π}, where 0 < λ₁< ⋯ < λ_r < 1.Then the value of λ₁ + ⋯ + λ_r is

(a) 1/4

(b) -2

(c) 1

(d) 1/2

Solution:

Given that f(θ) = (sin θ + cos θ)² + (sin θ – cos θ)⁴

= (sin² θ + cos² θ) + 2 sin θ cos θ + ((sin θ – cos θ)²)²

= 1 + sin 2θ + (sin² θ + cos² θ – 2sin θ cos θ)²

= 1 + sin 2θ + (1 – sin 2θ)²

= 1 + sin 2θ + 1+ sin² 2 θ – 2sin 2θ

= 2 + sin² 2 θ – sin 2θ

= sin² 2 θ – sin 2θ + 2

= (sin 2θ – ½)² + 7/4

Given that f has a local minimum at θ.

θ ∈ [0, π]

2 θ ∈ [0, 2 π]

f(θ) min. when sin 2 θ = ½

So, 2 θ = π/6, 5π/6

θ = π/12, 5π/12

λ₁ = 1/12 and λ₂ = 5/12

λ₁ + λ₂ = 6/12 = 1/2

Hence option d is the answer.

Question 8: The number of distinct solutions of equation (5/4)cos²2x + cos⁴x + sin⁴x + cos⁶x + sin⁶x = 2 in the interval [0, 2π] is

(a) 2

(b) 8

(c) 4

(d) 1

Solution:

(5/4)cos²2x + cos⁴x + sin⁴x + cos⁶x + sin⁶x = 2

=> (5/4)cos²2x + (cos²x)² + (sin²x)² + (cos²x)³ + (sin²x)³ = 2

=> (5/4)cos²2x + (cos²x + sin²x)² – 2 sin²x cos²x + (cos²x + sin²x)(sin⁴x + cos⁴x – sin²x cos²x) = 2

=> (5/4) cos²2x + 1 – 2 sin²x cos²x + (sin⁴x+ cos⁴x – sin²x cos²x) = 2

=> (5/4) cos²2x + 1 – 2 sin²x cos²x + (1 – 3 sin²x cos²x) = 2

=> (5/4) cos²2x – 5 sin²x cos²x = 0

(5/4) – (5/4)sin²2x – 5sin²x.cos²x = 0

=> (5/4) – (5/4)sin²2x – (5/4)sin²2x = 0

=> sin²2x = ½

=> sin 2x = ± 1/√2

So x = π/8, 3π/8, 5π/8, 7π/8, 9π/8, 11π/8, 13π/8, 15π/8

Number of solutions in [0, 2π] is 8.

Hence option b is the answer.

Question 9: For what and only what values of α lying between 0 and π is the inequality sin α cos³α > sin³α cos α valid?

(a) α ∈ (0, π/4)

(b) α ∈ (0, π/2)

(c) α ∈ (π/4, π/2)

(d) none of these

Solution:

sin α cos³α > sin³α cos α

=> sin α cos³α – sin³α cos α > 0

=> sin α cos α(cos²α – sin²α) >0 ..(i)

We know cos²x – sin²x = cos 2x.

So (i) becomes

(½)2sin α cos α cos 2α > 0

=> (½) sin 2α cos 2α > 0

=> (¼) sin 4α > 0

=> sin 4α > 0

=> 4α ∈ (0, π) (given that 0<α<π)

=> α ∈ (0, π/4)

Hence option a is the answer.

Question 10: The solution of the equation tan θ. tan 2 θ = 1 is

(a) nπ + 5π/12

(b) nπ – π/12

(c) 2nπ ± π/4

(d) nπ ± π/6

Solution:

We know tan 2θ = 2 tan θ/(1 – tan²θ)

Given tan θ. tan 2θ = 1

=> tan θ.2 tan θ/(1 – tan²θ) = 1

=> 2 tan²θ = 1 – tan²θ

=> 3 tan²θ = 1

=> tan²θ = 1/3

=> tan θ = 1/√3

=> θ = nπ ± π/6

Hence option d is the answer.

Question 11: If sin θ = 3 sin (θ + 2α), then the value of tan (θ + α) + 2 tan α is

(a) 3

(b) 1

(c) 2

(d) 0

Solution:

Given that sin θ = 3 sin (θ + 2α)

=> sin (θ + α – α) = 3 sin (θ + α + α)

=> sin (θ + α) cos α – cos(θ + α) sin α = 3 sin (θ + α) cos α + 3 cos(θ + α) sin α

=> -2 sin (θ + α) cos α = 4 cos(θ + α) sin α

=>-sin (θ + α)/cos (θ + α) = 2 sin α/cos α

=> -tan (θ + α) = 2 tan α

=> tan (θ + α) + 2 tan α = 0

Hence option d is the answer.

Question 12: The value of tan 6⁰ tan 42⁰ tan 66⁰ tan 78⁰ is

(a) 1

(b) 4

(c) 2

(d) 0

Solution:

tan 6⁰ tan 42⁰ tan 66⁰ tan 78⁰ = (sin 6⁰ sin 42⁰ sin 66⁰ sin 78⁰)/(cos 6⁰ cos 42⁰ cos 66⁰ cos 78⁰)

= (cos 60 – cos 72)(cos 36 – cos 120)/(cos 60 + cos 72)(cos 36 + cos 120)

= (½ – sin 18)(cos 36 + ½)/ (½ + sin 18)(cos 36 – ½)

= [(½ – (√5 – 1)/4) ((√5+1)/4 + ½ )]/[(½ + (√5 – 1)/4) ((√5+1)/4 – ½ )]

= (3 – √5)(3 + √5)/(√5 +1)(√5 -1)

= (9-5)/(5-1)

= 4/4

= 1

Hence option a is the answer.

Question 13: The value of √3 cosec 20⁰ – sec 20⁰ is equal to

(a) -6

(b) 4

(c) 1

(d) 0

Solution:

√3 cosec 20⁰ – sec 20⁰ = √3/sin 20⁰ – 1/cos 20⁰

= (√3 cos 20⁰ – sin 20⁰)/sin 20⁰ cos 20⁰

= 4[(√3/2) cos 20⁰ – ½ sin 20⁰)/2 sin 20⁰ cos 20⁰

= 4[(sin 60⁰ cos 20⁰ – cos 60⁰ sin 20⁰)/sin 20⁰]

= 4[ (sin (60⁰-20⁰))/sin 40⁰]

= 4 sin 40⁰/sin 40⁰

= 4

Hence option b is the answer.

Question 14: The value of the expression (1 – 4 sin 10⁰ sin 70⁰)/2 sin 10⁰ is

(a) -6

(b) 4

(c) 1

(d) 0

Solution:

(1 – 4 sin 10⁰ sin 70⁰)/2 sin 10⁰ = (1 – 2(2 sin 10⁰ sin 70⁰))/2 sin 10⁰

We know sin (90 – θ) = cos θ

And cos A – cos B = 2 sin (A+B)/2 sin (A-B)/2

(1 – 2(2 sin 10⁰ sin 70⁰))/2 sin 10⁰ = [1 – 2(cos 60⁰ – cos 80⁰)]/2 sin (90⁰ – 80⁰)

= (1- 2 cos 60⁰ + 2 cos 80⁰)/2 cos 80⁰

= (1 – 2(½) + 2 cos 80⁰)/2 cos 80⁰

= 2 cos 80⁰/2 cos 80⁰

= 1

Hence option c is the answer.

Question 15: Let a, b, c be three non zero real numbers such that the equation √3a cos x + 2b sin x = c, x belongs to [-π/2, π/2] has two distinct roots α and β with α + β = π/3. Then the value of b/a is

(a) 1

(b) 4/3

(c) 1/2

(d) 0

Solution:

Given that √3a cos x + 2b sin x = c

Divide by a, we get

√3 cos x + (2b/a) sin x = c/a …(i)

Since α and β are the roots of (i)

√3 cos α + (2b/a) sin α = c/a …(ii)

√3 cos β + (2b/a) sin β = c/a …(iii)

Subtract (iii) from (ii)

=> √3 (cos α – cos β) + (2b/a) (sin α – sin β) = 0

We know cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2

Also sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2

=> √3 (-2 sin (α+β)/2 sin (α-β)/2) + (2b/a) (2 cos (α+β)/2 sin (α-β)/2) = 0

Given α + β = π/3

=> √3 (-2 sin π/6 sin (α-β)/2) + (2b/a) (2 cos π/6 sin (α-β)/2) = 0

=> √3 (-2 ×½ × sin (α-β)/2) + (2b/a) (2×√3/2 sin (α-β)/2) = 0

=> √3 ( sin (α-β)/2) = (2b/a) (√3 sin (α-β)/2)

=> 1 = 2b/a

=> b/a = 1/2

Hence option c is the answer.

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