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

Explanation:

tan θ = tan α

∴ θ = nπ + α 

Concept:

• The domain of a function is the set of values that we are allowed to plug into our function.
•  The range is the set of all possible values that the function will give when we give in the domain as input.

Calculation:

The above graph is of the function cosecx

Since \(cosec x = \frac{1}{sinx}\), the domain of the cosec function is the set {x : x ∈ R and x ≠ nπ, n ∈ Z} or R - {x : x = nπ, n ∈ Z} and 

the range set is the set {y : y ∈ R and y ≥ 1 or y ≤ -1} 

i.e., the set R - (-1,1).

It means that y = cosec x assumes all real values except -1 < y < 1 and is not defined for the integral multiple of π.

Hence, the correct answer is option 2).

Explanation:

Given:

⇒ \(\rm \frac{x}{\cos θ}=\frac{y}{\cos \left(\frac{2\pi}{3}-θ\right)}=\frac{z}{\cos\left(\frac{2\pi}{3}+θ\right)}\) = k (say)

⇒ x = kcosθ

⇒ y = k \(cos\frac{2\pi }{3} -θ \)

⇒ z = k \(cos\frac{2\pi }{3} +θ \)

Now, x + y + z = k [cosθ + cos \((\frac{2\pi }{3} -θ) + cos(\frac{2\pi }{3} +θ)\)]

= k [cosθ + 2cos\(\frac{2\pi }{3}cosθ \)]

= k[cosθ + 2 (\(-\frac{1}{2}) cosθ \)]

= k[cosθ- cosθ ] =0

⇒ x + y + z =0

∴ Option (b) is correct.

Explanation:

⇒ \(\frac{cos17°- sin17°}{cos17°+ sin17°} = \frac{cos17°(1-\frac{sin 17°}{Cos17°})}{cos17°(1+\frac{sin 17°}{Cos17°})}\) 

= \(\frac{1-\tan17°}{1+\tan17°}\) 

= \(\frac{tan 45° - tan17°}{1 + tan 45°. tan17°}\)

= tan 28° = Cot 62°

∴ the Correct answer is Option 4.

Explanation:

Given

cot2x. cot3x = 1

⇒ cos2x. cos3x – sin2x.sin3x= 0

⇒ cos(2x+3x) = cos5x = 0    0 < x < π 

Hence 5x = \(\frac{\pi }{2}, \frac{3\pi }{2}, \frac{5\pi }{2}, \frac{7\pi }{2}, \frac{9\pi }{2}\)

⇒ x = \(\frac{\pi }{10}, \frac{3\pi }{10}, \frac{5\pi }{10}, \frac{7\pi }{10}, \frac{9\pi }{10}\)

Thus, only five solutions are possible

∴ the correct answer is option C

Concept:

a² - b² = (a - b) (a + b)

sec x = 1/cos x and cosec x = 1/sin x

a³ + b³ = (a + b) (a² + b² - ab)

Calculation:

 \(\frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} - {\rm{co}}{{\rm{s}}^2}{\rm{A}}} \right)}}{{{\rm{cosA}}\left( {{\rm{secA}} - {\rm{cosecA}}} \right)\left( {{\rm{si}}{{\rm{n}}^3}{\rm{A}} + {\rm{co}}{{\rm{s}}^3}{\rm{A}}} \right)}}\) 

⇒ \( \frac{{\left( {{\rm{1}} - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{1}{{{\rm{cosA}}}} - \frac{1}{{{\rm{sinA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{si}}{{\rm{n}}^2}{\rm{A}} + {\rm{co}}{{\rm{s}}^2}{\rm{A}} - {\rm{sinAcosA}}} \right)}}\)

⇒ \( \frac{{\left( {1 - {\rm{sinAcosA}}} \right)\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {{\rm{sinA}} - {\rm{cosA}}} \right)}}{{{\rm{cosA}}\left[ {\frac{{{\rm{sinA}} - {\rm{cosA}}}}{{{\rm{sinA}}.{\rm{cosA}}}}} \right]\left( {{\rm{sinA}} + {\rm{cosA}}} \right)\left( {1 - {\rm{sinAcosA}}} \right)}}\)

⇒ \(\frac{sinA - cosA}{cosA[\frac{sinA - cosA}{sinA.cosA}]}\)

⇒ \(\frac{(sinA - cosA)\times sinA.cosA}{cosA[sinA - cosA]}\)

⇒ \(\frac{ sinA.cosA}{cosA}\)

⇒ sin A

∴ The correct answer is option (1).

Concept:

cos 2x = cos² x – sin² x = 2 cos² x - 1 = 1 – 2 sin² x

Calculation:

cos 4x

= cos 2(2x)

= 2 cos² 2x – 1                                         (∵cos 2x = 2 cos² x – 1)

= 2 (1 – 2 sin² x)² – 1                              (∵ cos 2x = 1 – 2 sin² x)

= 2 [1 – 4 sin² x + 4 sin⁴ x] – 1               [∵ (a – b)² = a² – 2ab + b²]

= 2 - 8 sin² x + 8 sin⁴ x – 1

= 1 - 8 sin² x + 8 sin⁴ x

Concept:

If sin θ = sin α then θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.

Calculation:

Given: 4 sin 3x = 2

⇒ sin 3x = ½

As we know that, sin (π/6) = ½

⇒ sin 3x = sin (π/6)

As we know that, if sin θ = sin α then

θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.

⇒ 3x = nπ + (- 1)ⁿ × (π/6), where n ∈ Z.

Concept:

General solution of some standard trigonometric equations:

Calculation:

Given: cos x = 1

⇒ cos x = cos 0

As we know, If cos θ = cos α then θ = 2nπ ± α

Therefore, x = 2nπ ± 0 = 2nπ 

Hence, the general solution of cos x = 1 is x = 2nπ, n ∈ Z.

Concept:

sin2x + cos2x = 1

Calculations:

Given, 4sin2x -  2 cos2x = 2

⇒2sin2x -  cos2x = 1

⇒ 2sin2x -  cos2x = sin2x + cos2x

⇒ sin2x = 2 cos2x

⇒tan2 x = 2

⇒ tan x = √2

Concept:

\(\rm \sin \theta = \frac{Perpendicular}{Hypotenuse}\)

\(\rm \cos \theta = \frac{Base}{Hypotenuse} \)

sin2 θ + cos2 θ = 1 

Calculation:

7 sinθ + 24 cosθ = 25

Dividing by 25 on both the sides, we get

\(\rm \frac{7}{25}\)sinθ + \(\rm \frac{24}{25}\)cosθ = 1      ....(i)

We know that,

sin² θ + cos² θ = 1  

sin θ.sin θ + cos θ.cos θ = 1      ....(ii)

On comparing equ (i) and (ii)

sin θ = \(\rm \frac{7}{25}\) 

cos θ = \(\rm \frac{24}{25}\)

Now, (sinθ + cosθ) 

= \(\rm \frac{7}{25}\)+ \(\rm \frac{24}{25}\)

= \(\rm \frac{31}{25}\)

Concept:

\(\rm cot (A + B) = \dfrac{cot A cot B - 1}{cot A + cot B}\)

Calculation:

Given,

(cot 2x ∙ cot 4x) - (cot 4x ∙ cot 6x) - (cot 6x ∙ cot 2x)

⇒ (cot 2x ∙ cot 4x) - cot 6x [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) - cot (2x + 4x) [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) -  \((\rm \dfrac{cot 2x .cot 4x - 1}{cot 2x + cot 4x})\)× [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) - (cot 2x ∙ cot 4 x - 1)

⇒ (cot 2x ∙ cot 4x) - (cot 2x ∙ cot 4x) + 1

⇒ 1

Concept:

If tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

Calculation:

Given: 8 tan(2x) – 5 = 3

⇒ 8 tan(2x) = 8

⇒ tan 2x = 1

As we know that, tan (π/4) = 1

⇒ tan 2x = tan (π/4)

As we know that, if tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

⇒ 2x = nπ + (π/4), where n ∈ Z.

Concept:

cos (A + B) = cosA. cosB - sinA. sin B

Calculation:

Given: cos 2B = 3 sin² A   .... (1)

and  3sin 2A = 2 sin 2B     .... (2)

Divide (1) by (2),

\(\frac{cos(2B)}{2sin(2B)}=\frac{3sin^2(A)}{3sin(2A)}\) .... (3)

We know that,

sin 2A = 2 sin A∙ cos A

Hence, equation (3) becomes,

\(\frac{cos(2B)}{2sin(2B)}=\frac{sin^2(A)}{2sin(A)cos(A)}\)

⇒ \(\frac{cos(2B)}{sin(2B)}=\frac{sin(A)}{cos(A)}\)

⇒ cos (2B) cos (A) = sin (2B) sin (A)

⇒ cos (2B) cos (A) - sin (2B) sin (A) = 0

From above concept,

cos (A + 2B) = 0

⇒ (A + 2B) = cos^-1 (0)

⇒ (A + 2B) =  π/2

Concept:

2sinx cos x = sin 2x

sin (90° + x) = cos x

Calculation:

Given, 2 sin 75° cos 75°

= sin [2 (75°)]

= sin (150°)

= sin (90° + 60°)

= cos 60°

= \(\rm \dfrac 1 2\)

Hence, the value of 2 sin 75° cos 75° is \(\rm \dfrac 1 2\)