Direct Ratio MCQ Quiz in मल्याळम - Objective Question with Answer for Direct Ratio - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 26, 2025

നേടുക Direct Ratio ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Direct Ratio MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Direct Ratio MCQ Objective Questions

Top Direct Ratio MCQ Objective Questions

Direct Ratio Question 1:

If cotθ = 4/3, then evaluate: (secθ(1 + cot2θ)(cosec2θ - cot2θ)) / cosec3θ.

  1. 3/4
  2. 4/5
  3. 4/3
  4. 3/5

Answer (Detailed Solution Below)

Option 1 : 3/4

Direct Ratio Question 1 Detailed Solution

Calculation:

cosec2θ - cot2θ = 1, 

cosec2θ = 1 + cot2θ 

So,

(secθ(1 + cot2θ)(cosec2θ - cot2θ)) / cosec3θ

⇒ (sec θ × cosec2θ × 1) / cosec3θ

⇒ (sec θ) / (cosec θ)

⇒ (1/cos θ) / (1/sinθ)

⇒ sinθ / cosθ

⇒ tan θ = 1/cot θ = 1/(4/3) = 3/4

∴ The corrct answer is option (1).

Direct Ratio Question 2:

If \(\rm tan A = {3 \over 8},\) then the value of \({3 \sin A + 2 \cos A} \over 3 \sin A - 2 \cos A\) is:

  1. \(- {13 \over25}\)
  2. \(- {25 \over7}\)
  3. \({25 \over8}\)
  4. \({13 \over21}\)

Answer (Detailed Solution Below)

Option 2 : \(- {25 \over7}\)

Direct Ratio Question 2 Detailed Solution

Given:

\(\rm tan A = {3 \over 8},\)

Concept used:

TanA = SinA / CosA

Calculation:

\({3 \sin A + 2 \cos A} \over 3 \sin A - 2 \cos A\)

Dividing by CosA we get,

3 TanA + 2 / 3 TanA - 2

Now putting the value of Tan A we get,

⇒ \( (3 × {3\over8}+ 2)\over (3 × {3\over8}- 2)\)

⇒ \(- {25 \over7}\)

∴ The  correct option is 2

Direct Ratio Question 3:

If tan x = \(\frac{7}{5}\), then the value of \(\frac{9 \sin x - \frac{42}{5}\cos x}{15 \sin x + 21 \cos x}\) is:

  1. 0
  2. 1
  3. 0.1
  4. 0.5

Answer (Detailed Solution Below)

Option 3 : 0.1

Direct Ratio Question 3 Detailed Solution

Given

tan x = 7/5

Formula used

tan x = P/B

sin x = P/H

cos x = B/H

Calculation

tan x = 7/5 = P/B

H2 = P2 + B2

H2 = 72 + 52

H2 = 49 + 25 = 74

h = √74.

\(\frac{9 \sin x - \frac{42}{5}\cos x}{15 \sin x + 21 \cos x}\)

⇒ (9 × 7/√74 - 42/5 × 5/√74)/(15 × 7/√74 + 21 × 5/√74)

⇒ (21/√74)/(210/√74)

⇒ 21/210

⇒ 0.1

The value is 0.1

Direct Ratio Question 4:

If acot θ = b, then what will be the value of \(\frac{bcosθ - asinθ}{bcosθ + asinθ}\)?

  1. \(\frac{b^2 + a^2}{b^2 - a^2}\)
  2. b2 + a2
  3. \(\frac{b^2 - a^2}{b^2 + a^2}\)
  4. 0

Answer (Detailed Solution Below)

Option 3 : \(\frac{b^2 - a^2}{b^2 + a^2}\)

Direct Ratio Question 4 Detailed Solution

Given

a cot θ = b

Calculation

\(\frac{bcosθ - asinθ}{bcosθ + asinθ}\)

Dividing the whole equation by cosθ, we get:

⇒ (b - asinθ/cosθ)/(b + asinθ/cosθ)

⇒ (b - atanθ)/(b + atanθ)

⇒ (b - a × a/b)/(b + a × a/b) 

⇒ (b2 - a2)/(b2 + a2) 

The value of the expression is  (b2 - a2)/(b2 + a2) .

Direct Ratio Question 5:

If Cos A = \({{9} \over 41}\), find cot A. 

  1. \({{9} \over 40}\)
  2. \({{41} \over 40}\)
  3. \({{40} \over 9}\)
  4. \({{9} \over 41}\)

Answer (Detailed Solution Below)

Option 1 : \({{9} \over 40}\)

Direct Ratio Question 5 Detailed Solution

Given:

Cos A = 9/41 

Formula used:

 cot A = cos A/sin A

Cos2 A + sin2 A = 1

Calculation:

Cos2 A + sin2 A = 1

⇒ ( 9/41)2 + sin2 A = 1

⇒  sin2 A = 1 - 81/1681

 ⇒ sin2 A = 1600/1681

⇒  sin A = 40/41

 cot A = cos A/sin A

 ⇒ cot A = 9/40

∴ Option 1 is the correct answer.

Direct Ratio Question 6:

If sin θ = \(1 \over 2\), then the value of (3cos θ - 4 cos3 θ) is:

  1. 0
  2. 1
  3. 2
  4. -1

Answer (Detailed Solution Below)

Option 1 : 0

Direct Ratio Question 6 Detailed Solution

Given:

sin θ = \(1 \over 2\)

Concept used:

Trigo

Calculation:

sin θ = \(1 \over 2\)

⇒ sinθ = sin 30°

So, θ = 30°

3cos θ - 4 cos3 θ

⇒ 3 × \(\sqrt3 \over 2\) - 4 × \(3\sqrt3 \over 8\)

⇒ \({3\sqrt3 \over 2} - \frac{3\sqrt3}{2}\)

⇒ 0

∴ The required answer is 0.

Direct Ratio Question 7:

If tan α = 6, then sec α equals to:

  1. \(\sqrt{7}\)
  2. \(\sqrt{5}\)
  3. \(\sqrt{37}\)
  4. \(\sqrt{35}\)

Answer (Detailed Solution Below)

Option 3 : \(\sqrt{37}\)

Direct Ratio Question 7 Detailed Solution

Given:

Tan α = 6

Concept used:

Tan θ = P/B

Sec θ = H/B

Where, P = perpendicular; B = base; H = hypotenuse

Formula used:

Pythagorean theorem:

H2 = P2 + B2

Where, P = perpendicular; B = base; H = hypotenuse

Calculation:

Tan α = 6/1 = P/B

Using Pythagorean theorem:

⇒ H2 = P2 + B2

⇒ H2 = (6)2 + (1)2

⇒ H = √37

Sec α = H/B = √37

∴ The correct answer is √37. 

Direct Ratio Question 8:

If 4 tan θ = 3, then \(\frac{4 \sin θ-\cos θ+1}{4 \sin θ+\cos θ-1}=\) _________.

  1. \(\frac{14}{11}\)
  2. \(\frac{12}{11}\)
  3. \(\frac{10}{11}\)
  4. \(\frac{13}{11}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{13}{11}\)

Direct Ratio Question 8 Detailed Solution

Given:

4 tan θ = 3

Calculation:

F3 SSC Arbaz 24-05-2023 Himanshu D1

Applying Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 32 + 42

AC = 5

sinθ = 3/5

cosθ =4/5

\(\frac{4 \sin θ-\cos θ+1}{4 \sin θ+\cos θ-1}\)

Put the above values

⇒ \(\frac{4 \times 3/5-4/5+1}{4 \times 3/5+4/5-1}\)

⇒ \(\frac{12/5-4/5+1}{12/5+4/5-1}\) = \(\frac{(12-4+5)/5}{(12 +4 - 5)/5}\)

⇒ \(\frac{13/5}{11/5}\) = \(\frac{13}{11}\) 

Option 4 is the correct answer.

Direct Ratio Question 9:

If 7 tan θ = 3, and θ is an acute angle, then \(\frac{5 \sin \theta - \cos \theta}{5 \sin \theta + 2 \cos \theta}\) is equal to:

  1. \(\quad \frac{8}{29}\)
  2. \(\quad \frac{11}{29}\)
  3. \(\quad \frac{7}{29}\)
  4. \(\quad \frac{9}{29}\)

Answer (Detailed Solution Below)

Option 1 : \(\quad \frac{8}{29}\)

Direct Ratio Question 9 Detailed Solution

Given:

7tan θ = 3

Calculation:

\(\frac{5 \sin \theta - \cos \theta}{5 \sin \theta + 2 \cos \theta}\)

Divide by 'cos θ' in both numerator and denominator,

⇒ \(\frac{5 (\sin \theta/ \cos \theta) -1}{5 (\sin \theta/ \cos \theta) + 2 }\)

⇒ \(\frac{5\ tan \theta -1}{5 \ tan \theta + 2 }\)

⇒ [5 × 3/7 - 1] / [5 × 3/7 + 2]

⇒ [15/7 - 1] / [15/7 + 2]

⇒ [(15 - 7)/7] / [(15 + 14)/7]

⇒ [8/7] / [29/7]

⇒ 8/29

∴ The correct answer is option (1).

Direct Ratio Question 10:

Evaluate the value of \((cosec 56^{\circ} \ cos 34^{\circ} - cos 59^{\circ} \ cosec 31^{\circ})\).

  1. 1
  2. 2
  3. -1
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Direct Ratio Question 10 Detailed Solution

Given:

The expression to evaluate is: \((cosec 56^{\circ} \cdot \cos 34^{\circ} - cos 59^{\circ} \cdot cosec 31^{\circ})\)

Formula used:

The cosecant function is defined as: \(cosec x = \dfrac{1}{\sin x}\)

Properties used:

\(cosec x = \dfrac{1}{\sin x}\)

\(cos x\) is the cosine of angle x.

Step 1: Substitute the value of cosecant using the identity \(cosec x = \dfrac{1}{\sin x}\) into the expression:

We now have: \(\dfrac{1}{\sin 56^{\circ}} \cdot \cos 34^{\circ} - \cos 59^{\circ} \cdot \dfrac{1}{\sin 31^{\circ}}\)

Step 2: The expression becomes:

\(\dfrac{\cos 34^{\circ}}{\sin 56^{\circ}} - \dfrac{\cos 59^{\circ}}{\sin 31^{\circ}}\)

Step 2: Apply the identity \(cos(90^{\circ} - \theta) = sin \theta\) to both terms:

\(\dfrac{\cos (90^{\circ} - 56^{\circ})}{\sin 56^{\circ}} - \dfrac{\cos 59^{\circ}}{\sin (90^{\circ} - 59^{\circ})}\)

Step 3: Simplify the expression using the identity:

\(\dfrac{\sin 56^{\circ}}{\sin 56^{\circ}} - \dfrac{\cos 59^{\circ}}{\cos 59^{\circ}}\)

Step 4: Simplify the terms:

\(1 - 1\)

∴ The value of the expression is 0

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