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