Chapter 8 Introduction to Trigonometry- MCQ Online Test 2 Class 10 Maths
1. Find the value of tan15°tan20°tan7°tan75°?
a. 0
b. 2
c. 1
d. 3
2. Find the value of tan (55°−θ) − cot (35°+ θ)?
a. 0
b. 3
c. 2
d. 5
3. If cos 9α = sin α, then find the value of α?
a. 3°
b. 5°
c. 8°
d. 9°
4. If tan θ =√3, find the value of sec θ?
a. 3
b. 2
c. 5
d. 6
5. 
a. sinα
b. sin β
c. cos α
d. cosβ
6. If Cos A+Cos2 A=1, then find the value of Sin A+Sin2 A=1?
a. 2
b. 3
c. 1
d. 0
7. If sin A + 2 cos A = 1, then find the value of 2 sin A – cos A?
a. 5
b. 3
c. 2
d. 1
8. If sin A + Sin2 A = 1 then find the value of Cos2 A+Cos2 A?
a. 3
b. 2
c. 5
d. 1
9. If x cos A = 1 and tan A = y, then find the value of x²− y²?
a. 2
b. 1
c. 3
d. 5
10. If sinθ + cosθ = √2cosθ then find the value of cos θ – sin θ?
a. √2sinθ
b. √3sinθ
c. √5sinθ
d. √12sinθ
Chapter - 8 Introduction to Trigonometry Quiz-3 | Math Class 10th
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Question 1 of 10
1. Question
Find the value of tan 15° tan 20° tan 7° tan 75° ?
Correct
Given: tan 15° tan 20° tan 70° tan 75°
= tan 15° tan 20° tan (90°− 20° ) tan (90° −15°)
= tan 15° tan 20° cot 20° cot 15°
= (tan 15° cot 15°) (tan 20° cot 20° )
= 1 x 1 = 1Incorrect
Given: tan 15° tan 20° tan 70° tan 75°
= tan 15° tan 20° tan (90°− 20° ) tan (90° −15°)
= tan 15° tan 20° cot 20° cot 15°
= (tan 15° cot 15°) (tan 20° cot 20° )
= 1 x 1 = 1 -
Question 2 of 10
2. Question
Find the value of tan (55°−θ) − cot (35°+ θ)?
Correct
Given: tan (55° −θ) −cot (35° +θ)
cot (90° −55° +θ) −cot (35° +θ)
cot (35° +θ) −cot (35° +θ) = 0Incorrect
Given: tan (55° −θ) −cot (35° +θ)
cot (90° −55° +θ) −cot (35° +θ)
cot (35° +θ) −cot (35° +θ) = 0 -
Question 3 of 10
3. Question
If cos 9α = sin α, then find the value of α?
Correct
Given: cos9α = sinα
L.H.S =
cos9α = sin (90°−9α)
R.H.S = sinα
On comparing L.H.S and R.H.S we get
90−9α = α
90 = 10α
α = 9°Incorrect
Given: cos9α = sinα
L.H.S =
cos9α = sin (90°−9α)
R.H.S = sinα
On comparing L.H.S and R.H.S we get
90−9α = α
90 = 10α
α = 9° -
Question 4 of 10
4. Question
If tan θ =√3, find the value of sec θ?
Correct
Incorrect
-
Question 5 of 10
5. Question
Correct
Incorrect
-
Question 6 of 10
6. Question
If CosA +Cos2 A=1, then find the value of Sin A +Sin2 A = 1?
Correct
Given: Cos A +Cos2 A=1
Cos A = 1−cos2A
Cos A= sin2 A
Squaring both sides, we get
Cos2 A = Sin4 A
1−Sin2 A = Sin4 A
Sin2A + Sin2A = 1Incorrect
Given: Cos A +Cos2 A=1
Cos A = 1−cos2A
Cos A= sin2 A
Squaring both sides, we get
Cos2 A = Sin4 A
1−Sin2 A = Sin4 A
Sin2A + Sin2A = 1 -
Question 7 of 10
7. Question
If sin A + 2 cos A = 1, then find the value of 2 sin A – cos A?
Correct
Given: sin A + 2cos A = 1
(Sin A+2Cos A)2 =1
Sin2 A+ 4 Cos2 A+2 SinA. 2CosA=1
1-Cos2 A+4(1-Sin2 A)+4 Sin A. Cos A=1
1-cos2A+4-4 Sin2A+ 4sinA. CosA =1
-Cos2A+4-4 sin2A+4SinA.CosA=0
Multiplying by – sign
Cos2A-4+(2sinA)2-2.(2SinA).CosA =0
(2SinA)2+Cos2 A-2.(2sinA).CosA = 4
(2 sin A-Cos A)2 = 22
2 sin A – Cos A = 2Incorrect
Given: sin A + 2cos A = 1
(Sin A+2Cos A)2 =1
Sin2 A+ 4 Cos2 A+2 SinA. 2CosA=1
1-Cos2 A+4(1-Sin2 A)+4 Sin A. Cos A=1
1-cos2A+4-4 Sin2A+ 4sinA. CosA =1
-Cos2A+4-4 sin2A+4SinA.CosA=0
Multiplying by – sign
Cos2A-4+(2sinA)2-2.(2SinA).CosA =0
(2SinA)2+Cos2 A-2.(2sinA).CosA = 4
(2 sin A-Cos A)2 = 22
2 sin A – Cos A = 2 -
Question 8 of 10
8. Question
If sin A + Sin2 A = 1 then find the value of Cos2 A+Cos4 A?
Correct
Given: Sin A+Sin2 A = 1
Sin A+Sin2 A=1
Sin A=1-Sin2 A
Sin A=Cos2 A
Sin2 A=Cos4 A
1-Cos2 A=(Cos4 A)
Cos4 A+Cos2 A=1Incorrect
Given: Sin A+Sin2 A = 1
Sin A+Sin2 A=1
Sin A=1-Sin2 A
Sin A=Cos2 A
Sin2 A=Cos4 A
1-Cos2 A=(Cos4 A)
Cos4 A+Cos2 A=1 -
Question 9 of 10
9. Question
If x cos A = 1 and tan A = y, then find the value of x²− y²?
Correct
Incorrect
-
Question 10 of 10
10. Question
If sinθ + cosθ = √2 cosθ then find the value of cos θ – sin θ?
Correct
Given: sinθ + cosθ = √2cosθ
Squaring both sides, we get
sin²θ+cos²θ+2sinθcosθ = 2cos²θ
cos²θ − 2sinθ cos θ = sin2θ
cos²θ−2sinθ cosθ+sin2θ = 2sin² θ
(cos θ –sin θ) ² = 2sin2θ
cosθ−sinθ = √2sinθIncorrect
Given: sinθ + cosθ = √2cosθ
Squaring both sides, we get
sin²θ+cos²θ+2sinθcosθ = 2cos²θ
cos²θ − 2sinθ cos θ = sin2θ
cos²θ−2sinθ cosθ+sin2θ = 2sin² θ
(cos θ –sin θ) ² = 2sin2θ
cosθ−sinθ = √2sinθ