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

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).

Given:

$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{A}=\frac{3}{8},$

Concept used:

TanA = SinA / CosA

Calculation:

$\frac{3\sin A+2\cos A}{3\sin A-2\cos A}$

Dividing by CosA we get,

3 TanA + 2 / 3 TanA - 2

Now putting the value of Tan A we get,

⇒ $\frac{(3\times \frac{3}{8}+2)}{(3\times \frac{3}{8}-2)}$

⇒ $-\frac{25}{7}$

∴ The  correct option is 2

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

Given

a cot θ = b

Calculation

$\frac{bcos\theta -asin\theta }{bcos\theta +asin\theta }$

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) .

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.

Given:

sin θ = $\frac{1}{2}$

Concept used:

Calculation:

sin θ = $\frac{1}{2}$

⇒ sinθ = sin 30°

So, θ = 30°

3cos θ - 4 cos3 θ

⇒ 3 × $\frac{\sqrt{3}}{2}$ - 4 × $\frac{3\sqrt{3}}{8}$

⇒ $\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}$

⇒ 0

∴ The required answer is 0.

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. 

Given:

4 tan θ = 3

Calculation:

Applying Pythagoras Theorem

AC² = AB² + BC²

AC² = 3² + 4²

AC = 5

sinθ = 3/5

cosθ =4/5

$\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}$

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.

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,

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

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

⇒ [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).

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: $cosecx={\displaystyle \frac{1}{\sin x}}$

Properties used:

$cosecx={\displaystyle \frac{1}{\sin x}}$

$cosx$ is the cosine of angle x.

Step 1: Substitute the value of cosecant using the identity $cosecx={\displaystyle \frac{1}{\sin x}}$ into the expression:

We now have: ${\displaystyle \frac{1}{\sin {56}^{\circ }}}\cdot \cos {34}^{\circ }-\cos {59}^{\circ }\cdot {\displaystyle \frac{1}{\sin {31}^{\circ }}}$

Step 2: The expression becomes:

${\displaystyle \frac{\cos {34}^{\circ }}{\sin {56}^{\circ }}}-{\displaystyle \frac{\cos {59}^{\circ }}{\sin {31}^{\circ }}}$

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

${\displaystyle \frac{\cos ({90}^{\circ }-{56}^{\circ })}{\sin {56}^{\circ }}}-{\displaystyle \frac{\cos {59}^{\circ }}{\sin ({90}^{\circ }-{59}^{\circ })}}$

Step 3: Simplify the expression using the identity:

${\displaystyle \frac{\sin {56}^{\circ }}{\sin {56}^{\circ }}}-{\displaystyle \frac{\cos {59}^{\circ }}{\cos {59}^{\circ }}}$

Step 4: Simplify the terms:

$1-1$

∴ The value of the expression is 0