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) sin^{2}x

d) cosec(x2)

View Answer

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 sin

^{2}x the part laying in the –y axis comes in the positive y axis.

2. What is the value of tanθ?

a) √(1 + cos^{2}θ)/cosθ

b) √(1 – cos^{2}θ)/cosθ

c) (√(1 – cos^{2}θ))cosθ

d) √(1 – cos^{2}θ)+cosθ

View Answer

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 – x

^{2})

By definition, AB/OA = √(1 – x

^{2})/x

= √(1 – cos

^{2}θ)/cosθ.

3. What will be the value of (sinx + cosecx)^{2} + (cosx + secx)^{2} ?

a) ≥ 0

b) ≤ 0

c) ≤ 1

d) ≥ 1

View Answer

Explanation: The given expression in LHS is,

sin

^{2}x + cosec

^{2}x + 2 + cos

^{2}x + sec

^{2}x + 2

4 + (sin

^{2}x + cos

^{2}x) +(1 + tan

^{2}x) + (1 + cot

^{2}x)

= 7 + (tan

^{2}x + cot

^{2}x)

= 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

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

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).

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

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

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) (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

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)

Similarly, tan (B/2) = (c – a)/(c + a) (cot(C – A)/2).

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

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

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

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.

12. The given graph is for which equation?

a) |sinx|

b) |cotx|

c) |sinx|

d) Unpredictable

View Answer

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.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 11**.

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