Trigonometric Equations 2 - Class 11 Maths MCQ

This set of Class 11 Maths Chapter 3 Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Equations – 2”.

1. Which of the following will form the following graph?

a) sinx + cosx
b) 2sin(x/2)
c) sin²x
d) cosec(x2)
View Answer

Answer: c
Explanation:

This is the curve of sinx.
It is given that the graph in the question does not lay in the –y axis.
So, when we plot the graph of sin²x the part laying in the –y axis comes in the positive y axis.

2. 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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).

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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
Explanation: (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

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

9. The given graph is for which equation?

a) y = (x – 1)(x – 2)
b) y = (x + 1)(x + 2)
c) y = |(x – 1)(x – 2)|
d) y = |(x + 1)(x + 2)|
View Answer

Answer: c
Explanation: As the critical points are 1and 2 so the graph will form in this range only.
Clearly, as the equation is enclosed in modulus so,
The graph will lie in the positive y direction and will approach infinity.

10. The given graph is for which equation?

a) y = |(x + 1)(x – 3)|
b) y = |(x + 1)(x + 3)|
c) y = |(x – 1)(x – 3)|
d) y = -|(x + 1)(x – 3)|
View Answer

Answer: a
Explanation: As, the critical points are -1 and 3,
Therefore, the curve will lie from (-1, 3)
So, this curve will cut the positive y axis at y = 3 and reach to the maximum height and come to point 3 at positive x axis.
Generally it attains a maximum height of 4 units in positive y axis.
As, the equation is enclosed by modulus so, it will lie only to upper y axis.

11. The given graph is for which equation?

a) cosecx
b) cotx
c) secx
d) cosx
View Answer

Answer: c
Explanation: There are 2 curves.

The blue curve is the graph of y = cosx
The red curve is the graph for y = secx
As, secx is reciprocal of cosx, so,
The graph of secx is the reciprocal of the graph of cosx.
So, the point at which cosx attain maximum height i.e. 1 and -1 in positive and negative y axis is the point of origin of this graph that approaches to infinity.

Answer: b
Explanation: The above graph is for |cotx|

The blue curve is for y = cotx and red line is for |cotx|
As, half the part of graph is below x axis is negative so, due to modulus it moves to above x axis.
The graph approaches to infinity.

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