250+ TOP MCQs on Trigonometric Equations & Answers | Class 11 Maths

Mathematics Multiple Choice Questions and Answers for Class 11 on “Trigonometric Equations – 2”.
1. What is the value of tanθ?
a) √(1 + cos²θ)/cosθ
b) √(1 – cos²θ)/cosθ
c) (√(1 – cos²θ))cosθ
d) √(1 – cos²θ)+cosθ
View Answer
Answer: b
Clarification: If AOB is a righted angled triangle with ∠AOB = θ and ∠BAO = 90°
Also, consider OA = x and OB = 1
By definition, cosθ = OA/OB = x/1 = x
So, AB = √(1 – x²)
By definition, AB/OA = √(1 – x²)/x
= √(1 – cos²θ)/cosθ.
3. What will be the value of (sinx + cosecx)² + (cosx + secx)² ?
a) ≥ 0
b) ≤ 0
c) ≤ 1
d) ≥ 1
View Answer
Answer: a
Clarification: The given expression in LHS is,
sin²x + cosec²x + 2 + cos²x + sec²x + 2
4 + (sin²x + cos²x) +(1 + tan²x) + (1 + cot²x)
= 7 + (tan²x + cot²x)
= 7 + (tanx – cotx)² + 2 which is ≥ 0.
4. What will be the value of dy/dx = (x + 2y + 3)/(2x + 3y + 4)?
a) -2π – 9
b) 2π – 9
c) -2π + 9
d) 2π + 9
View Answer
Answer: c
Clarification: We know, sec x = sec(π – x)
So, sec 6 = sec(π – 9)
= sec(2π + 9)
= sec(3π – 9)
= sec(-π – 9)
= sec(-2π – 9)
= sec(-3π – 9)
So, sec^-1(sin 9) = sin^-1(sin (-2π + 9))
= -2π + 9.
5. Which one is correct for Napier’s Analogy?
a) tan (C/2) = (a + b)/(a – b) (tan(A – B)/2)
b) tan (C/2) = (a – b)/(a + b) (cot(A – B)/2)
c) tan (C/2) = (a – b)/(a + b) (cot(A + B)/2)
d) tan (C/2) = (a + b)/(a – b) (tan(A + B)/2)
View Answer
Answer: b
Clarification: According to the law of sines, in any triangle ABC,
a/sinA = b/sinB = c/sinC
So, a/b = sinA/sinB
(a + b)/(a – b) = (sinA + sinB)/( sinA – sinB)
=> (a + b)/(a – b) = (2 sin((a + b)/2) cos((a – b)/2))/ (2 sin((a + b)/2) sin((a – b)/2))
=> (a + b)/(a – b) = (tan(A + B)/2)/(tan(A – B)/2)
=> (tan(A + B)/2) = (a + b)/(a – b) (tan(A – B)/2)
=> (tan(π/2 + C/2)) = (a + b)/(a – b) (tan(A – B)/2)
=> cot (C/2) = (a + b)/(a – b) (tan(A – B)/2)
=> tan (C/2) = (a – b)/(a + b) (cot(A – B)/2).
6. What is the value of (1 + cotA)(1 + cotB) for isosceles right triangle ABC right angled at A?
a) 0
b) 1
c) 2
d) Data inadequate
View Answer
Answer: c
Clarification: (1 + cotA)(1 + cotB)
Simplifying the above equation,
= 1 + cotA + cotB + cotAcotB
Putting the values of A and B, we get,
= 1 + cot90 + cot45 + cot90cot45
= 1 + 0 + 1 + 0
= 2.
7. Which one is correct for Napier’s Analogy?
a) tan (A/2) = (b – c)/(b + c) (tan(B + C)/2)
b) tan (A/2) = (b – c)/(b + c) (cot(B – C)/2)
c) tan (A/2) = (b + c)/(b – c) (cot(B – C)/2)
d) tan (A) = (b – c)/(b + c) (tan(B – C)/2)
View Answer
Answer: b
Clarification: According to the law of sines, in any triangle ABC,
a/sinA = b/sinB = c/sinC
So, a/b = sinA/sinB
(a + b)/(a – b) = (sinA + sinB)/( sinA – sinB)
=> (a + b)/(a – b) = (2 sin((a + b)/2) cos((a – b)/2))/ (2 sin((a + b)/2) sin((a – b)/2))
=> (a + b)/(a – b) = (tan(A + B)/2)/(tan(A – B)/2)
=> (tan(A + B)/2) = (a + b)/(a – b) (tan(A – B)/2)
=> (tan(π/2 + C/2)) = (a + b)/(a – b) (tan(A – B)/2)
=> cot (C/2) = (a + b)/(a – b) (tan(A – B)/2)
=> tan (C/2) = (a – b)/(a + b) (tan(A – B)/2)
Similarly, tan (A/2) = (b – c)/(b + c) (cot(B – C)/2).
8. Which one is correct for Napier’s Analogy?
a) tan (B/2) = (c – a)/(c + a) (cot(C + A)/2)
b) tan (B/2) = (c – a)/(c + a) (cot(C – A)/2)
c) tan (B/2) = (c + a)/(c – a) (cot(C – A)/2)
d) tan (B) = (c – a)/(c + a) (cot(C – A)/2)
View Answer
Answer: b
Clarification: According to the law of sines, in any triangle ABC,
a/sinA = b/sinB = c/sinC
So, a/b = sinA/sinB
(a + b)/(a – b) = (sinA + sinB)/( sinA – sinB)
=> (a + b)/(a – b) = (2 sin((a + b)/2) cos((a – b)/2))/ (2 sin((a + b)/2) sin((a – b)/2))
=> (a + b)/(a – b) = (tan(A + B)/2)/(tan(A – B)/2)
=> (tan(A + B)/2) = (a + b)/(a – b) (tan(A – B)/2)
=> (tan(π/2 + C/2)) = (a + b)/(a – b) (tan(A – B)/2)
=> cot (C/2) = (a + b)/(a – b) (tan(A – B)/2)
=> tan (C/2) = (a – b)/(a + b) (cot(A – B)/2)
Similarly, tan (B/2) = (c – a)/(c + a) (cot(C – A)/2).