MCQs on Trigonometry – Set 1

¹.π⁄2

² π⁄3

³.π⁄4

⁴ π⁄6

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

d. √(₁₀₊₂√₅) ⁄₄

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

a. 2

b. 4

c. 3

d. 1

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

Explanation

tan 1⁰. tan 2⁰ ……. tan 89⁰

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

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

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

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

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

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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