MCQs on Trigonometry – Set 2
Q1. Solution of the eq
sin²θ+3cos²θ=4 is
a. ηπ±π∕6
b.2ηπ±π∕6
c.ηπ±π∕3
d.2ηπ±π∕3
Answer: Option ‘a’ is correct
Explanation:
7sin²θ+3cos²θ=4
⇒7sin²θ+3(1-sin²θ)=4
⇒7sin²θ+3-3sin²θ=4
⇒4sin²θ+3=4
⇒4sin²θ=1
⇒sin²θ=1∕4
⇒sinθ=±1∕2
⇒θ=±π∕6
Q2. If tan (πcosθ)=cot(πsinθ), then cos(θ-π∕4) is equale to;
a. ±1∕3√2
b. 1∕2√2
c.±1∕2√2
d. -1∕2√2
Answer: option ’c’ is correct
Explanation;
tan (πcosθ)=cot(πsinθ)
⇒sin(πcosθ)∕cos(πcosθ) = cos(πsinθ)∕sin(πsinθ)
⇒sin(πcosθ).sin(πsinθ)=cos(πsinθ).cos(πcosθ)
⇒sin(πcosθ).sin(πsinθ)-cos(πsinθ).cos(πcosθ)
⇒cos(πcosθ+πsinθ)=0
⇒πcosθ+πsinθ=±π∕2
cosθ+sinθ=±1∕2
Multiply both sides by 1∕√2
⇒ 1∕√2cosθ+1∕√2sinθ=±1∕2√2
since cosπ∕4=1∕√2=sinπ∕4
⇒cosπ∕4.cosθ+sinπ∕4.π∕4.sinθ=±1∕2√2
⇒cos(θ-π∕4)=±1∕2√2
Q3. Value of sin (deg 1) ∕ sin (rad 1) is
a. >1
b. <1
c. =1
d. =π∕180
Answer: option ‘b’ is correct
Explanation:
since deg 1 < rad 1, sin (deg 1)<sin(rad 1)
⇒sin (deg 1) ∕ sin(rad 1) <1
Q4. If sin (x+y) ∕ sin (x-y) = a+b ∕ a-b
then value of tan x ∕ tan y is ;
a. a∕b
b. b∕a
c. ab
d. (a-b)(a+b)
Answer: option ’a’ is correct
Explanation;
Use componendo dividendo theorem
Q5. If sin θ= sin α, then
a. θ+α ∕2 is any odd multiple of π∕2 and θ-α∕2 is any multiple of π
b. θ+α ∕2 is any even multiple of π∕2 and θ-α∕2 is any multiple of π
c. θ+α ∕2 is any multiple of π∕2 and θ-α∕2 is any odd multiple of π
d. θ+α ∕2 is any multiple of π∕2 and θ-α∕2 is any even multiple of π
Answer: option ’a’ is correct
Explanation:
sin θ= sin α
⇒ sin θ – sin α=0
⇒ 2cos θ+α ∕2. sin θ-α ∕2
now one of these is 0 i-e
cos θ+α ∕2 =0 or sin θ-α ∕2=0
⇒ θ+α ∕2 is any odd multiple of π∕2 and θ-α∕2 is any multiple of π
Q6. If α+β+γ=2π
then the value of tanα∕2+tanβ∕2+tanγ∕2
is equale to;
a. tanα/2.tanβ/2.tanγ/2
b. tanα∕2tanβ∕2+tanβ∕2.tanγ∕2+tanα∕2.tanγ∕2
c. 1-tanα∕2.tanβ∕2.tanγ∕2
d. 1+tanα∕2.tanβ∕2.tanγ∕2
Answer: option ’a’ is correct
Explanation
Given
α+β+γ=2π
⇒α∕2+β∕2+γ∕2=π
⇒tan (α∕2+β∕2)=tan(π-γ∕2)
⇒tanα∕2+tanβ∕2 ∕ 1-tanα∕2.tanβ∕2 = -tanγ∕2
⇒tanα∕2+tanβ∕2+tanγ∕2=tanα∕2.tanβ∕2.tanγ∕2
Q7. If 1+sin x + sin² x +sin ³ x +………..+∞ = -4+2√3 for 0<x<π ,
then the value of x is;
a. π/6 or 2π/3
b. π/4 or 2π/5
c. π/3 or π/6
d. π/3 or 2π/3
Answer: option ’ d’ is correct
Explanation:
1+sin x + sin² x +sin ³ x +………..+∞ = -4+2√3
⇒1/ 1- sin x = -4+2√3
⇒sin x = √3/2
⇒ x = π/3 or 2π/3
Q8. If tan θ+sin θ =m and tan θ – sin θ=n
then;
a. m²-n² =16mn
b. m²+n² =16mn
c. (m²-n²)² =16mn
d. (m²+n²)² =16mn
Answer: option ’ c’ is correct
Explanation:
You can prove it by putting values of m and n
Q9, The value of tan 31°.tan 32°.tan 33° ……………tan 59° is equale to
a. -1
b. 1
c. 0
d. 2
Answer: option ’ c’ is correct
Explanation
tan 31°.tan 32°.tan 33° ……………tan 59°
=(tan 31°. tan58°) X (tan 32°. tan 58°)………x(tan 44°. tan 46°)x (tan 45°)
=(tan 31° . tan(90°-31°)) x (tan 32°. tan (90°-32°)) x ………. x (tan 44° .tan(90°-44°)) x tan 45°
= (tan 31° cot 31°) x (tan 32° . cot 32°) x ……..x(tan44° . cot 44°) x 1 = 1
Q10.The numbers tan[-11π/6] . tan[21π/4] . cot [283π/6] are in
a. A.P.
b. C.P.
c. H.P.
d, G.P.
Answer: Option ’b’ is correct
Explanation:
tan[-11π/6] . tan[21π/4] . cot [283π/6]
or
tan[-330] . tan[945] . cot [8490]
or
-tan[360-30] . tan[2×360+225] . cot [23×360+210]
or
tan[30] . tan[225] . cot [210]
or
tan[30] . tan[π+45] . cot [π+30]
or
tan[30] . tan[45] . cot [30]
or
1/√3.1.√3
which ar in G.P. with common ratio of √3