Test: Trigonometric Functions - 2 - JEE MCQ

# Test: Trigonometric Functions - 2 - JEE MCQ

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## 25 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Trigonometric Functions - 2

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Test: Trigonometric Functions - 2 - Question 1

### Angles of triangle are in A.P. If the number of degrees in the smallest to the number of radians in the largest is as 60:π, then the smallest angle is

Detailed Solution for Test: Trigonometric Functions - 2 - Question 1

Let the angles of the triangle are (a - d), a, (a + d)
Now, in a triangle, the sum of all angles equal to 180 degree
⇒ (a - d) + a + (a + d) = 180
⇒ a - d + a + a + d = 180
⇒ 3a = 180
⇒ a = 180/3
⇒ a = 60
Now, the angle are (60 - d), 60, (60 + d)
Now, 60 - d is the least and 60 + d is the greatest angle.
Now, (60 + d)° = {(60 + d) * (π/180)}c
Given that, number of radians in the greatest angle/number of degrees in the least one = π/60
⇒ {(60 + d) * (π/180)}c /(60 - d) = π/60
⇒ (60 - d)/{(60 + d) * (π/180)}c = 60/π
⇒ 180(60 - d)/{(60 + d) * π} = 60/π
⇒ 180(60 - d)/(60 + d) = 60
⇒ (60 - d)/(60 + d) = 60/180
⇒ (60 - d)/(60 + d) = 1/3
⇒ 3(60 - d) = (60 + d)
⇒ 180 - 3d = 60 + d
⇒ 180 - 60 = 3d + d
⇒ 4d = 120
⇒ d = 120/4
⇒ d = 30

Test: Trigonometric Functions - 2 - Question 2

### If x, y, z are perpendiculars drawn from the vertices of a triangle having sides a, b and c, then bx/c + cy/a + az/b =

Detailed Solution for Test: Trigonometric Functions - 2 - Question 2

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Test: Trigonometric Functions - 2 - Question 3

### The largest value of sin θ cos ⁡θ is

Detailed Solution for Test: Trigonometric Functions - 2 - Question 3

1/root 2 ; root 2 is largest value of sin theta and cos thetha

Test: Trigonometric Functions - 2 - Question 4

If tan θ + sec θ = √3, 0 < π , then θ is equal to

Detailed Solution for Test: Trigonometric Functions - 2 - Question 4

Test: Trigonometric Functions - 2 - Question 5

If sin (120°−α) = sin(120°−β) and 0 <α,β<π, then all values of α and β are given by

Detailed Solution for Test: Trigonometric Functions - 2 - Question 5

sin(120 – α) = sin(120 – β), 0 < α and β < π
⇒ either 120 – α = 120 – β ⇒ α = β]
or 120 – α = 180 – (120 – β)
∴ 120 – α = 60 + β
∴ α + β = 60° = (π/3)
∴ α = β or α + β = (π/3).

Test: Trigonometric Functions - 2 - Question 6

The equation (cos π – 1) x2 + cos π) x + sin π = 0, where x is a variable, has real roots. Then the interval of π may be any one of the following:

Detailed Solution for Test: Trigonometric Functions - 2 - Question 6

Discriminant of the given equation = (cos p) 2 − 4(cosp − 1) sin p
= cos2p + 4 (1 − cosp) sin p ≥ 0, if p ∈ (0, π)
[∵ cos2p ≥ 0, 0 ≤ 1 − cosp ≤ 2 and sin p > 0 for all p ∈ (0, π)]

Test: Trigonometric Functions - 2 - Question 7

In a triangle ABC, a = 13, b = 14, c = 15, then r1 =

Detailed Solution for Test: Trigonometric Functions - 2 - Question 7

Test: Trigonometric Functions - 2 - Question 8

If sinθ+cosecθ = 2, then sin2θ+cosec2θ =

Detailed Solution for Test: Trigonometric Functions - 2 - Question 8

sinθ+cscθ=2
⇒ sinθ+1/sinθ=2
⇒ sin2θ−2sinθ+1=0
⇒ (sinθ−1)2 = 0
⇒ sinθ=1
⇒ sin2θ + csc2θ
= sin2θ + 1/sin2θ
= 1+1
= 2

Test: Trigonometric Functions - 2 - Question 9

The value of tan 75 - cot 75 is equal to

Detailed Solution for Test: Trigonometric Functions - 2 - Question 9

tan75° - cot75°
= sin75°/cos75° - cos75°/sin75°
= 2(sin2 75° - cos2 75°)/2sin75°cos75°
⇒ -2cos150°/sin150°
= -2cot150°
= -2cot(180-30)°  = 2cot30°
= 2(3)1/2

Test: Trigonometric Functions - 2 - Question 10

In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals

Detailed Solution for Test: Trigonometric Functions - 2 - Question 10

Test: Trigonometric Functions - 2 - Question 11

If the angles of at triangle are in the ratio 1 : 2 : 3, then the sides are in the ratio

Detailed Solution for Test: Trigonometric Functions - 2 - Question 11

Let the angles of a triangle are x° ,2x°. 3x°.acordingly:-

x°+2x° +3x°=180°

6x°=180°

or. x=30°

Angles are 30° , 60°. and. 90°.

Let a. , b and c units be the length of the sides of the triangle. Applying

sine rule

a/sin30° = b /sin60° = c/sin90°

2.a/1= 2b/√3. = c/1

a= c/2 , b= (c.√3)/2.

a : b : c = c/2. : c√3/2. : c. = 1/2. : √3/2 : 1. or 1 : √3. : 2. Answer.

Test: Trigonometric Functions - 2 - Question 12

In a triangle ABC right angled at C, tan A and tan B satisfy the equation

Detailed Solution for Test: Trigonometric Functions - 2 - Question 12

Test: Trigonometric Functions - 2 - Question 13

Let the angles A, B, C of ΔABC be in A.P. and let b: c:: √3 : √2, then the angle A is

Detailed Solution for Test: Trigonometric Functions - 2 - Question 13

As the angles A, B, C of ∆ ABC are in AP
∴ Let A = x – d, B = x, C = x + d
But A + B + C = 180° (∠ Sum prop. of ∆)
∴ x – d + x + x + d = 180°
⇒ 3x = 180° ⇒ x = 60° ∴ ∠B = 60°
Now by sine law in ∆ ABC, we have
b/sin B = c/sin C ⇒ sin B/sin C
⇒ √3/√2 = sin 60°/sin C [using b : c = √3 : √2 and ∠B = 60°]
⇒ √3/√2 = √3/2 sin C ⇒ sin C = 1/√2 = sin 45°
∴ ∠C = 45° ⇒ ∠A = 180° - (∠B + ∠C)
= 180° - (60° + 45°) = 75°

Test: Trigonometric Functions - 2 - Question 14

cot θ = sin 2θ(θ ≠ nπ, n integer) if θ equals

Test: Trigonometric Functions - 2 - Question 15

If A = sin2θ+cos4θ then for all real values of θ

Detailed Solution for Test: Trigonometric Functions - 2 - Question 15

Test: Trigonometric Functions - 2 - Question 16

The general solution of tan 3x = 1 is (n ∈ I)

Test: Trigonometric Functions - 2 - Question 17

The value of sin

Detailed Solution for Test: Trigonometric Functions - 2 - Question 17

sinθ = sin(π−θ)
sin(9/14π) = sin(π−9/14π) = sin5/14π
(sin11/14)π = (sin3/14)π
(sin 13/14)π = (sinπ/14)
= (sin2 π/14) (sin2 3/14)π (sin2 5/14)π
= [(sinπ/14) (sin3/14)π (sin5/14)π]2
= 1/[(4sin2π)/7][sin2/7 π soc 2/7π ⋅ cos(π−4/7π)]2
= 1/16sin2 π/7[1/4(sin28/7)π]
= 1/64{sin2(π+π/7)sin2 π/7}
= 1/64{sin2(π/7)sin2(π/7)}
= 1/64

Test: Trigonometric Functions - 2 - Question 18

If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ (= PR), then the angle P is

Test: Trigonometric Functions - 2 - Question 19

The maximum value of sin  in the interval  is attained at

Detailed Solution for Test: Trigonometric Functions - 2 - Question 19

cos(x+π/6) + sin(x+π/6)
√2[1/√2 cos(x+π/6) + 1/√2 sin(x+π/6)]
= √2 cos[x + π/6 - π/4]
= √2 cos[x - π/12]
For maximum value x = π/12

Test: Trigonometric Functions - 2 - Question 20

The solution of tan 2θ tan θ = 1 is

Detailed Solution for Test: Trigonometric Functions - 2 - Question 20

Test: Trigonometric Functions - 2 - Question 21

sin (180+φ) sin(180−φ) cosec2 φ

Detailed Solution for Test: Trigonometric Functions - 2 - Question 21

Test: Trigonometric Functions - 2 - Question 22

The general solution of the equation sin x + cos x = 2 is

Detailed Solution for Test: Trigonometric Functions - 2 - Question 22

sinx+cosx=2—(1)sinx+cosx=2—(1)

Divide by 2–√2on both LHSandRHS=>sinx∗12√+cosx∗12√=22√=2–√—(2)LHSandRHS=>sinx∗12+cosx∗12=22=2—(2)

Since sin45=cos45=12√sin45=cos45=12, the equation(2)(2) can be written assinxcos45+cosxsin45=2–√—(3)sinxcos45+cosxsin45=2—(3)

SincesinAcosB+cosAsinB=sin(A+B)sinAcosB+cosAsinB=sin(A+B), equation (3)(3)can be written as sin(x+45)=2–√—(4)sin(x+45)=2—(4)

The known facts are:

• The value for sinAsinAis always between−1−1and11
• 2–√=1.4142=1.414, which is greater than 1

From the above facts, LHS of the equation (4) is always between -1 and 1 whereas RHS is > 1. This is not possible and hence six+cosxsix+cosx can never be equal to 2.

Test: Trigonometric Functions - 2 - Question 23

Which of the following is not possible ?

Detailed Solution for Test: Trigonometric Functions - 2 - Question 23

CosA value can vary between -1 and +1..here any value of "t" except "0" will fall out of this range

Test: Trigonometric Functions - 2 - Question 24

If (1 + tan θ) (1 + tan φ) = 2, then θ + φ =

Detailed Solution for Test: Trigonometric Functions - 2 - Question 24

Test: Trigonometric Functions - 2 - Question 25

If x = y cos 2π/3 = z cos 4π/3, then xy + yz +zx

Detailed Solution for Test: Trigonometric Functions - 2 - Question 25

Thus, the value of xy + yz + zx is 0.

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