Trigonometric Identities MCQ Quiz - Objective Question with Answer for Trigonometric Identities - Download Free PDF

Given:

3sin θ + 5cos θ = 5

Concept:

sin² θ + cos² θ = 1

a² + b² + 2ab = (a + b)²

a2 + b2 - 2ab = (a - b)2

Calculation:

Let 5sin θ - 3cos θ = x    ....(1)

And,

3sin θ + 5cos θ = 5       ....(2)

Squaring in both the equation:

9sin² θ + 25cos² θ + 30sin θ. cos θ= 25                     ....(3)

25sin2 θ + 9cos2 θ - 30sin θ. cos θ  = x²               .....(4)

From equation (3) and equation (4)

⇒ 34 (sin2 θ + cos2 θ) = 25 + x2

⇒ 9 = x2

x = 3 and -3

∴ 5sin θ - 3cos θ = 3

Concept:

 \(sin\frac{x}{2}\) = \(\sqrt{{\frac{1-cosx}{2}}} \)

 \(cos\frac{x}{2}\) = \(\sqrt{{\frac{1+cosx}{2}}} \)

Calculation:

Given:

cos α and cos β are the roots of the equation  8x2 - 2x - 1 = 0

8x2 - 2x - 1 = 0

⇒ 8x2 - 4x + 2x - 1 = 0

⇒ 4x (2x - 1) + (2x - 1) = 0

⇒ (4x + 1) (2x - 1) = 0

⇒ x =  \(-\frac{1}{4}\)  and x = \(\frac{1}{2}\)

Since cos α and cos β are the roots of the equation,

∴ cos α =  \(-\frac{1}{4}\)  and cos β = \(\frac{1}{2}\)

⇒  \(cos\frac{α}{2}\) = \(\sqrt{{\frac{1+cosα}{2}}} \)

⇒ \(cos\frac{α}{2}=\sqrt{{\frac{1-\frac{1}{4}}{2}}} \)

⇒ \(cos\frac{α}{2}=\sqrt\frac{3}{8}=\frac{\sqrt3}{2\sqrt2}\)      ------(1)

and  \(cos\frac{β}{2}\) = \(\sqrt{{\frac{1+cosβ}{2}}} \)

⇒ \(cos\frac{β}{2}\) = \(\sqrt{{\frac{1+\frac{1}{2}}{2}}} \)

⇒ \(cos\frac{β}{2}\) = \(\frac{\sqrt3}{2} \)      ------(2)

Multiplying equations (1) and (2), we get

\(cos\frac{α}{2}cos\frac{β}{2}=\frac{3}{4\sqrt2}\)

∴​ the value of \(cos\frac{α}{2}cos\frac{β}{2}=\frac{3}{4\sqrt2}\)

Concept:

The opposite angles of the cyclic quadrilateral are supplementary.

\(\rm \angle A + \angle C = 180^{∘}\)

\(\rm \angle B + \angle D = 180^{∘}\)

cos(180^∘ - θ) = - cos θ 

cos(180∘ + θ) = - cos θ 

sin(90° - θ) = cosθ 

Calculator:

Statement 1

L.H.S

cos A + cos B

⇒ cos (180^∘ - C) + cos (180^∘ - D)

⇒ - cos C - cos D

⇒ - (cos C + cos D) = R.H.S

Statement 2

L.H.S

cos (π + A) + cos (π - B) + cos (π - C) - sin \(\rm \left(\frac{\pi}{2}-D\right) \)

⇒ - cos A - cos B - cos C - cos D

⇒ - cos (180∘ - C) - cos (180∘ - D) - cos C - cos D

⇒ cos C + cos D - cos C - cos D

⇒ 0 = R.H.S

∴ Both the statements are correct.

Given:

sin 3θ cosθ - cos 3θ sinθ = \(\frac{1}{2}\), 0 < θ < \(\frac{\pi}{2}\) 

Formula used:

sin A cosB - cos A sinB = sin(A - B)

Calculation:

Using formula, sin 3θ cosθ - cos 3θ sinθ = sin(3θ - θ) = sin(2θ)

⇒ sin(2θ) = 1/2

⇒ sin(2θ) = 1/2 =sin 30° [∵ 0 < θ < \(\frac{\pi}{2}\)]

⇒ 2θ = 30°

⇒ θ = 15°

Concept:

If A + B = 45° then

(1 + tan A)(1 + tan B) = 2

Calculation:

Given that,

(1 + tan θ) (1 + tan 9θ) = 2

We know that, if A + B = 45° then

(1 + tan A)(1 + tan B) = 2

⇒ θ + 9θ = π/4

⇒ 10θ = π/4

Hence, the required value

tan 10θ = tan(π/4)

∴  tan 10θ = 1

Additional Information 
∵ tan (A + B) = tan 45° = (tan A + tan B) / (1 – tan A tan B)
 
⇒ 1 – tan A tanB = tan A + tan B
 
⇒ 1 + 1 = 1 + tan A  + tan B + tan A tan B
 
⇒ (1 + tan A) (1 + tan B) = 2

Concept:

cosec² x – cot² x = 1

Calculation:

Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)^-1

⇒ cosec θ + cot θ = 1/q

As we know that, cosec² x – cot² x = 1

⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1

\(\frac1q \times p=1\)

Concept:

sin² x + cos² x = 1

Calculation:

Given: sin θ + cos θ = 7/5 

By, squaring both sides of the above equation we get,

⇒ (sin θ + cos θ)² = 49/25

⇒ sin² θ + cos² θ + 2sin θ.cos θ = 49/25

As we know that, sin2 x + cos2 x = 1

⇒ 1 + 2sin θcos θ = 49/25

⇒ 2sin θcos θ = 24/25

Concept:

I. sec² θ – tan² θ = 1

II. a² – b² = (a - b) (a + b)

Calculation:

Given:

tan θ + sec θ = 4     ...(1)

As we know that, sec² θ – tan² θ = 1

⇒ sec² θ – tan² θ = 1

⇒ (sec θ – tan θ) (sec θ + tan θ) = 1

By substituting the value of tan θ + sec θ = 4, in the above equation, we get

⇒ sec θ – tan θ = 1/4     ... (2)

Adding equation (1) and (2), we get

⇒ 2 sec θ = 17/4

⇒ sec θ = 17/8

cos θ = 8/17

Mistake PointsIt is tan θ which is equal to 15/8. The above two equations that we solved were:

tan θ + sec θ = 4

sec θ – tan θ = 1/4

Concept:

a2 - b2 = (a - b) (a + b)

sec2 x - tan2 x = 1

Calculation:

sec4 x - tan4 x

=(sec2 x - tan2 x) (sec2 x + tan2 x)          (∵ a2 - b2 = (a - b) (a + b))

= 1 × (1 + tan2 x + tan2 x)                                          (∵ sec2 x - tan2 x = 1)

= 1 + 2tan² x

Formula used

sin2A + cos2A = 1

1/sinA = cosecA

1/cosA = secA

Calculation

\(\rm sinA \left(1 + {sinA\over cosA} \right) + cosA \left(1 + {cosA\over sinA} \right)?\)

⇒ sinA (cosA + sinA)/cosA + cosA(sinA + cosA)/sinA

⇒ (sinA + cosA) [(sinA/cosA) + (cosA/sinA)]

⇒ (sinA + cosA) [(sin²A + cos²A)/(cosA.sinA)]

⇒ (sinA + cosA) [1/(cosA.sinA)]

⇒ 1/sinA + 1/cosA

⇒ secA + CosecA

The answer is secA + CosecA.

Given:

\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ 

Formula:

(a + b)² = a² + b² + 2ab

sin²θ + cos²θ = 1

Calculation:

\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ 

⇒ \(cos^3θ\over sin^3θ\) × sin²θ + \(sin^3θ\over cos^3θ\) × cos²θ + 2sinθ cosθ 

⇒ \(cos^3θ\over sinθ\) + \(sin^3θ\over cosθ\) + 2sinθ cosθ

⇒ \({sin^4θ+cos^4θ+2sin^2θ cos^2θ}\over{sinθ cosθ}\) 

⇒ \(({sin^2θ+cos^2θ})^2\over{sinθ cosθ}\) = \(1\over{sinθ cosθ}\)

⇒ \({1\over sinθ}\times{1\over cosθ}\) = cosecθ secθ (\(1\over{sinθ}\) = cosecθ, \(1\over{cosθ}\) = secθ) 

∴ \(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ = cosecθ.secθ

Concept:

• cosec x = 1/sin x
• cot x = cos x/ sin x
• cos² x + sin²x = 1

Calculation:

Here we have to find the value of \(\frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} - \sqrt {1 - {\rm{sinx}}} }}\)

By rationalizing the expression \(\frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} - \sqrt {1 - {\rm{sinx}}} }}\) we get

\(= \frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} - \sqrt {1 - {\rm{sinx}}} }}{\rm{\;}} \times \;\frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}\)

\( = \frac{{[\left( {1 + sinx + 1 - sinx} \right) + 2\sqrt {1 - si{n^2}x} }}{{\left( {1 + sinx - 1 + sinx} \right)}}\)

\( = \frac{{2 + 2{\rm{cosx}}}}{{2{\rm{\;sinx}}}}\)

∴ cosec x + cot x

Concept:

Trigonometry Formula

sec² θ = 1 + tan² θ

cosec2 θ = 1 + cot2 θ

cot θ = \(\rm \frac{1}{tan \theta }\)

sec θ = \(\rm \frac{1}{cos \theta }\)

cosec θ = \(\rm \frac{1}{sin \theta }\)

Calculation:

\(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\)

\(= \frac{sec^2θ}{cosec^2θ}-\left(\frac{1-tanθ}{1-\frac{1}{tan θ }}\right)^2\)

\(= \rm \frac{\frac{1}{cos^{2}θ}}{\frac{1}{sin^{2}θ}} - \left (\frac{1 - tanθ}{\frac{tan θ - 1}{tan θ }} \right )^{2}\)

\(\rm = \frac{sin^{2}θ }{cos^{2} θ } - \left (\frac{1 - tanθ}{\frac{- (1 - tan θ )}{tan θ }} \right )^{2}\)

= tan² θ - (-tan θ)²

= tan2 θ - tan2 θ 

= 0

Concept:

sec2 x - tan2 x = 1

Calculation:

Given sec x + tan x = 2     ....(i)

∵ sec² x - tan² x = 1

(sec x + tan x)(sec x - tan x) = 1

2(sec x - tan x) = 1

sec x - tan x = \(1\over 2\)              ....(ii)

Adding the equation (i) and (ii)

2 sec x = 2 + \(1\over 2\)

\(\rm{2\over \cos x} = {5\over 2}\)

cos x = \(4\over 5\)

Given:

√3 - 3√3tan2A = 3tan A - tan3A

Formula used:

tan 3A = (3tan A - tan3A)/(1 - 3tan²A)

Calculation:

√3 - 3√3tan2A = 3tan A - tan3A

⇒ √3(1 - 3tan2A) = 3tan A - tan3A

⇒ √3 = (3tan A - tan3A)/(1 - 3tan2A)

⇒ √3 = tan 3A 

⇒ tan 60° = tan 3A

⇒ 3A = 60° 

⇒ A = 60°/3 = 20° 

∴ The value of A is 20°.