Direct Ratio MCQ Quiz - Objective Question with Answer for Direct Ratio - Download Free PDF

Given:

sin θ = 3/5

cos θ = 4/5

Find the value of √((1 - tan²θ) / (cot²θ - 1))

Formula used:

tan θ = sin θ / cos θ

cot θ = 1 / tan θ

Calculation:

tan θ = sin θ / cos θ

⇒ tan θ = (3/5) / (4/5)

⇒ tan θ = 3/4

cot θ = 1 / tan θ

⇒ cot θ = 4/3

Now, calculate √((1 - tan²θ) / (cot²θ - 1))

tan²θ = (3/4)² = 9/16

cot²θ = (4/3)² = 16/9

⇒ √((1 - tan²θ) / (cot²θ - 1))

⇒ √((1 - 9/16) / (16/9 - 1))

⇒ √((16/16 - 9/16) / (16/9 - 9/9))

⇒ √((7/16) / (7/9))

⇒ √((7/16) × (9/7))

⇒ √(9/16)

⇒ 3/4

∴ The correct answer is option (1).

Given:

Formula used:

and

We need to find the value of:

Calculations:

First, find using :

Next, find . Since , we have:

Now substitute and into the expression:

⇒

Simplify the numerator:

Simplify the denominator:

Thus, the expression becomes:

⇒

⇒ Simplify the fraction:

Answer:

The value is .

Given:

8cotθ = 7

Concept Used:

Cotθ = 7/8

Cotθ = Base/Perpendicular

Calculation:

Hypotenuse = √(Perpendicular² + Base²)

⇒ √(8² + 7²)

⇒ √113

⇒ (1 + 8/√113)/(7/√113)

⇒ (8 + √113)/ 7   

∴ The value of  is (8 + √113)/ 7

Given:

secθ = √2

Formula Used:

secθ = 1/cosθ

cscθ = 1/sinθ

Calculation:

The value of cscθ is √2 .

Given:

tan A = 1/√10

Concept Used:

Sin A = Perpendicular/Hypotenuse

Cosec A = Hypotenuse/Perpendicular

tan A = Perpendicular/Base

Hypotenuse = √( Perpendicular² + Base²)

Calculation:

Hypotenuse = √( Perpendicular2 + Base2)

Hypotenuse = √( √102 + 12)

⇒ √ 11

sinA + cosecA

According to the concept,

1/√11 + √11/1

⇒ 12/√11

∴ The value of sinA + cosecA is 12/√11

Given : 

Calculation : 

We know,

⇒ Secθ = 4/3, tanθ = √7/3

⇒ tan2 θ + tan4 θ = 7/9 + 49/81

⇒ (63 + 49)/81

⇒ 112/81

∴ The correct answer is 112/81.

Given

  = 15

  = 16

Calculation

Sec α = cos β/15  ......(1)

sin β = 16 × Sec α  .....(2)

Putting the value of eq (1) in eq (2)

⇒ sin β = 16 × cos β/15

⇒ sin β/cos β = 16/15

⇒ P/B = 16/15

sin2β = P²/H²

H² = P² + B²

H2 = 256 + 225

H2 = 481

sin2β = 256/481

The value is 256/481.

Given:

b sin θ= a

Concept used:

sinθ = Perpendicular/ Hypotenuse

secθ = Hypotenuse/ Base

tanθ = Perpendicular/ Base

Calculation:

bsinθ = a

sinθ = a/b

So, Perpendicular = a &  Hypotenuse = b

(Hypotenuse)² = (Perpendicular)² + (Base)²

b² = a²+ Base²

base² = b²- a²

secθ = Hypotenuse/ base= b/

tanθ= perpendicular/ base= a/

So, secθ+ tanθ = b/+ (a/)

secθ + tanθ = (b+ a)/()

∵ (b + a) = √(b + a) × √(b + a)  

⇒ secθ + tanθ = / 

secθ+ tanθ= 

Given:

cos x = - (1/2)

Formula used:

Cos (180 + θ) = - cos θ 

Calculation:

cos x = - (1/2)

⇒ cos x = cos (180 + 60)

⇒ cos x = cos 240

⇒ x = 240°

Tan 240° = tan (180 + 60)

⇒ tan 60° = √3

∴ The correct answer is √3.

Given:

 = 10 ,  = 11

Calculation:

10 sinβ = 11 cosβ

⇒ tanβ = 11/10

cos²β = 100/221

The correct option 2

Formula used

sec θ + tan θ = 1/(sec θ - tan θ)

Calculation

If, sec θ + tan θ = 5, then:

sec θ - tan θ = 1/5

Adding the above two we get:

⇒ 2 sec θ = 5 + 1/5

⇒ sec θ = 1/2(5 + 1/5)

The value is 

Formula used:

Sec x = (1/cos x)

Calculation:

Sec x - cos x

⇒ (1/cos x) - cos x

⇒ (1 - cos² x)/cos x

⇒ sin² x/cos x

⇒ tan x sin x

∴ The correct answer is  tan x sin x.

Given:

Concept used:

(H)Hypotenuse² = (P)Perpendicular² + (B)Base²

Calculation:

sinθ = P/H

So, H² = P² + B²

⇒ 17² = 8² + B²

⇒ 289 = 64 + B2

⇒ 225 = B2

⇒ 15 = B

Now,

tanθ = P/B

So, tanθ =

∴ The required answer is .

 Shortcut Trick

sinθ = 8/17 = P/H, we know the triplet P = 8, B = 15, H = 17

Now, tanθ = P/B So, tanθ = 8/15

Given

tan x = -12/5

Concept

tan x = Perpendicular/base

Sin x = Perpendicular/hypotenuse

Calculation

tan x = -12/5 = Perpendicular/base

P = 12 and B = - 5

H² = P² + B²

H2 = 144 + 25

H2 = 169

H = 13

⇒ sin x − cot x

⇒ P/H + B/P   ...(since cot is also negative in second quadrant)

⇒ (P² + BH)PH

⇒ (144 + 65)/156

⇒ 209/156

The value is 209/156.