MCQs on Trigonometry – Set 1

1. The smallest value of θ satisfying the equation √₃ (cotθ+tanθ)=₄ is

¹.π⁄2

² π⁄3

³.π⁄4

⁴ π⁄6 

 
Answer – option 4
Explanation:

√₃ (cotθ+tanθ)=₄

cotθ+tanθ=₄ ⁄√₃

cos θ ⁄ sin θ + sinθ ⁄ cosθ = ₄ ⁄ √₃

cos²θ+sin²θ ⁄ sinθ cosθ = ₄ ⁄ √₃

2 ⁄ sin2θ = ₄ ⁄ √₃

sin2θ = √₃ ⁄ 2

2θ = π ⁄ ₃

θ = π ⁄ ₆

Q.2 Sin 18⁰ = ?
a. √₅ ₋ 1 ⁄ ₄
b. √₅₊₁ ⁄₄
c. √(₁₀₋₂√₅) ⁄₄
d. √(₁₀₊₂√₅) ⁄₄
Answer: Option  a is correct

Q 3. if sin x+sin²x=1 then cos⁸x+2cos⁶x+cos⁴x= ?
a. 0
b. -1
c.1/2
d. 1
Answer : option ‘d’ is correct
Explanation:

sin x+sin²x=1

sinx = 1-sin²x

=cos²x

Now putting sinx in place of cos²x in the second part i-e

cos⁸x+2cos⁶x+cos⁴x

=sin⁴x+2sin³x+sin²x

= (sin²x+sinx)²=(1)²=1


Q4. tan15⁰+tam75⁰=?

a. 2

b. 4

c. 3

d. 1

Answer: Option ‘b’ is correct
Explanation:

tan15+tan75= tan(45⁰-30⁰)+tan(45⁰+30⁰)

 

=tan45-tan30/1+tan45tan30 + tan45+tan30/1-tan45tan30

 

=₁₋₁ ⁄√³ ⁄ 1+ ₁ ⁄ √³ + ₁+₁ ⁄√³ ⁄ 1- ₁ ⁄ √³

 

= √₃₋₁ / √₃₊₁ + √₃+₁ / √₃-₁

 

=(√₃₋₁)² +(√₃₊₁ )² / (√₃-₁ )(√₃₊₁ )

 

=2(3+1) / 3-1

=8/2=4


Q5. tan 1⁰. tan 2⁰ ……. tan 89⁰ =?
a. 0
b. 1
c. -1
d. none of these
Answer : option ‘b’ is correct
Explanation

tan 1⁰. tan 2⁰ ……. tan 89⁰

=(tan1⁰ tan89⁰)(tan2⁰ tan88⁰)…….(tan44⁰ tan 46⁰) tan45⁰
=(tan1⁰ cot1⁰)(tan2⁰ cot2⁰)…….(tan44⁰ cot 44⁰) tan45⁰
=1


Q6. If cosα=cosβ and sinα=sinβ, then
a. cos α-β/2=0
b.cos α+β/2=0
c.sin α-β/2=0
d.sin α+β/2=0
Answer: option ‘d’ is correct
Explanation:
Given

cosα=cosβ and sinα=sinβ

=≻ cosα-cosβ=0 and sinα-sinβ=0

=≻ cosα-cosβ=0=sinα-sinβ

=≻ cosα-cosβ=sinα-sinβ=0

=≻ -2sin α+β⁄₂ sinα−β⁄₂=2cosα+β⁄₂ sinα−β⁄₂=0

=≻ sinα−β⁄₂=0


Q7. If A lies in the third quadrant and  3tan A -4=0 then 5sin 2A +3sin A +4 cos A=?
a.0
b. 5/6
c. 3/4
d. 2/3
Answer: option ‘a’ is correct
Explanation

Given

3tan A -4=0

=> tan A = 4/3

=> sin A = -4/5 and cos A =-3/5

=> sin 2A=2sin A cos A= 2( -4/5)(-3/5)

Given expression;

5(24/25)+3(-4/5)+4(-3/5)

=(24/5)-(12/5)-(12/5)=0


Q8. if secθ=√₂ and ₃π/₂<θ<₂π then ₁+tanθ+cosecθ / ₁+cotθ−cosecθ =?
a.−₁
b. ₀
c. ₁
d. none of these
Answer: option ‘a’ is correct
Explanation

if secθ=√₂ then cosθ=₁√₂

now sinθ=±√(1-cos²θ)= ±√(₁−₁ ⁄₂)=±₁ ⁄ √₂

but θ lies in third quadrant in which sinθ is negative

∴ sinθ=-₁ ⁄√₂

=> cosecθ=-√₂

now tanθ=sinθ/cosθ=−₁ ⁄ √₂ ×√₂ / ₁=−₁

=> cotθ=−₁

putting values in given expression

=> ₁+tanθ+cosecθ ⁄ ₁+ cotθ−cosecθ =₁−₁−√₂ ⁄ ₁−₁+√₂=−₁


Q9. If cotα=¹ ⁄ ₂ , α∈ (π, ₃π ⁄ ₂) and secβ= ₅ ⁄ ₃ , β∈ (π ⁄ ₂ , π)
then tan(α+β)=?
a. ₁ ⁄ ₁₁
b.− ₁ ⁄ ₁₁
c. ₂ ⁄ ₁₁
d. none of these

 

Answer: option ‘c’ is correct
Explanation

tan(α+β)= tanα+tanβ ⁄ ₁ − tanα tanβ

Given cosα= ₁ ⁄ ₂ => tanα= ₂

also

secα= −₅ ⁄ ₃

then

tanβ = √(sec²β − ₁)= ±√(₂₅ ⁄ ₉ − ₁)=±₄ ⁄ ₃

but β ∈ (π ⁄ ₂ , π) => tanβ=−₄ ⁄ ₃

putting value of tanα and tanβ in formula of

tan(α+β) = tanα+tanβ ⁄ ₁ − tanα.tanβ

= ₂− ₄ ⁄ ₃ ⁄ ₁− ₂ × (−₄ ⁄ ₃ )= ₆ − ₄ ⁄ ₃ + ₈ = ₂ ⁄ ₁₁

Q 10. The principal solution of cos x = – √₃ ⁄ ₂ are
a.₃π ⁄₆ , ₇π ⁄ ⁶
b. ₃π ⁄₆ , ₅π ⁄ ⁶
c. ₅π ⁄₆ , ₇π ⁄ ⁶
d. none of these
Answer: option ‘c’ is correct
Explanation

cos x is given as negative i-e ‘x’ lies in  2nd or 3rd quadrant !

Now cos x = – cos π ⁄ ₆ = cos (π−π ⁄ ₆)

or cos (π+π ⁄ ₆)= cos ₅π ⁄ ₆ or cos ₇π ⁄ ₆

=> x = ₅π ⁄ ₆ , ₇π ⁄ ₆

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