Content Writer
The trigonometric Function is a type of function that is used to express the relationship between the angles and sides of a triangle. It is an important concept used in NCERT Class 11 Mathematics, and for practice, students can try NCERT Solutions For Class 11 Mathematics Chapter 3: Trigonometric Functions.
- Trigonometric Functions are also called popular functions or circular functions.
- The functions are used to evaluate trigonometric values.
- Trigonometric identities and formulas are used to determine relationships between functions.
- Sine, cosine, tangent, cosecant, secant, and cotangent are six basic trigonometric functions.
- The functions can be expressed in terms of coordinates of x and y.
- Sine, cosine and tangent are three primary trigonometric functions.
- Other functions are derived from these functions using formulas.
Some formulas used to determine the trigonometric functions are as follows:
- \(\sin θ\) = \(\frac{\text{Opposite side}}{\text{Hypotenuse}}\)
- \(\cos θ\) = \(\frac{\text{Adjacent side}}{\text{Hypotenuse}}\)
- \(\tan θ\) = \(\frac{\text{Opposite side}}{\text{Adjacent side}}\)
- \(\cot θ\) = \(\frac{\text{Adjacent side}}{\text{Opposite Side}}\)
- \(\sec θ\) = \(\frac{\text{Hypotenuse}}{\text{Adjacent side}}\)
- \(cosec \;θ\) = \(\frac{\text{Hypotenuse}}{\text{Opposite Side}}\)
Trigonometric Functions MCQs
Ques: Calculate the value of cos75°.
- 1
- (√3 - 1) / 2√2
- (√3) / 2√2
- (√3 - 1) / 2
Click here for the answer
Ans: (a) (√3 - 1) / 2√2
Explanation: The formula for these types of functions include cos(A + B) = CosA.CosB - SinA.SinB.
⇒ In this the value of A = 30° and B = 45°
⇒ cos 75° = Cos(30° + 45°)
⇒ Cos30°.Cos45° - Sin30°.Sin45°
⇒ (√3/2) (1/√2) - (1/2) (1/√2)
⇒ 1/2√2 - √3/2√2
∴ the result is (√3 - 1) / 2√2
Ques: Calculate the value of cos 1° cos 2° cos 3° … cos 179°
- 1/√2
- 1
- 0
- –1
Click here for the answer
Ans: (c) 0
Explanation: cos 1° cos 2° cos 3° … cos 179°
⇒ cos 1° cos 2° cos 3° … cos 179°
⇒ cos 1° cos 2° cos 3° … cos 89° cos 90° cos 91° … cos 179°
⇒ cos 1° cos 2° cos 3° … cos 89° (0) cos 91° … cos 179°
∴ the result is 0 (since the value of cos 90° = 0)
Ques. The value of sin (30° + θ) – cos (30° – θ) is
- 2 cosθ
- 2 sinθ
- 1
- 0
Click here for the answer
Ans: (d) 0
Explanation: sin (30° + θ) – cos (30° – θ)
⇒ sin (30° + θ) – sin (90° -(30° – θ)) {since sin(90° – A) = cos A}
⇒ sin (45° + θ) – sin (30° + θ)
∴ the result is 0
Ques. Calculate the value of tan 1° tan 2° tan 3° … tan 89°
- 2
- 1
- ½
- 3
Click here for the answer
Ans: (b) 1
Explanation: tan 1° tan 2° tan 3° … tan 89°
⇒ [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]
⇒ [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°] × [tan 45°]
⇒ [(tan 1° × cot 1°) (tan 2° × cot 2°)…..(tan 44° × cot 44°)] × [tan 45°]
⇒ We know that, tan A × cot A =1 and tan 45° = 1
Hence, the equation becomes as: 1 × 1 × 1 × 1 × …× 1
∴ the result is 1 {As 1ⁿ = 1}
Ques: Calculate the value of cos 45o.
- 1/√2
- 2
- 1
- 0
Click here for the answer
Ans: (a) 1/√2
Explanation: cos 45o is equal to 1/√2
Ques. What is the value of the trigonometric function: cos(x–y).
- cosx.cosy–sinx.siny
- cos(x)
- cosx.cosy+sinx.siny
- 0
Click here for the answer
Ans: (c)cosx.cosy+sinx.siny
Explanation: cos(x-y) = cosx.cosy–sinx.siny
Ques: If the value of sec θ + tan θ = 64, then evaluate sec θ – tan θ.
- 1
- 6
- 7/64
- 1/64
Click here for the answer
Ans: (d) 1/64
Explanation: We know, sec2θ – tan2θ =1
⇒ (sec θ + tan θ)( sec θ – tan θ)=1
⇒ 64(sec θ – tan θ)=1
⇒ Or, sec θ – tan θ =1/64
So, the value of sec θ – tan θ =1/64
Ques. Determine the sign of sin 310°
- Positive
- Negative
- Both positive and negative
- Cannot be determined
Click here for the answer
Ans: (b) negative
Explanation: For the required angle, we first need to determine the coterminal angle which lies between 0° and 360°.
⇒ 270° < 300° < 360°
⇒ 310° lies in the fourth quadrant of the graph.
So, sin 300° is negative.
Ques: Suppose you have real values of x where cos θ = x + (1/x)
- θ is an acute angle
- θ is right angle
- No value of θ is possible
- θ is obtuse angle
Click here for the answer
Ans: (c) No value of θ is possible
Explanation: Since it is given that, cos θ = x + (1/x)
⇒ cos θ = (x2 + 1)/x
⇒ x2 + 1 = x cos θ
⇒ x2 – x cos θ + 1 = 0
⇒ Any real root of the equation is as follows: ax2 + bx + c = 0, b2 – 4ac ≥ 0.
⇒ (-cos θ)2 – 4 ≥ 0
⇒ cos2θ – 4 ≥ 0
⇒ cos2θ ≥ 4
⇒ cos θ ≥ ± 2
⇒ We know that -1 ≤ cos θ ≤ 1.
Hence, no value of θ is possible.
Ques. Determine the number of solutions of the equation tan x + sec x = 2 cos x which is lying in the interval [0, 2π]
- 0
- 1
- -2
- 2
Click here for the answer
Ans: (d) 2
Explanation: Since it is given that tan x + sec x = 2 cos x
⇒ (sin x/cos x) + (1/cos x) = 2 cos x
⇒ (sin x + 1)/cos x = 2 cos x
⇒ sin x + 1 = 2 cos2x
⇒ Using the identity sin2A + cos2A = 1,
⇒ sin x + 1 = 2(1 – sin2x)
⇒ sin x + 1 = 2 – 2 sin2x
⇒ 2 sin2x + sin x − 1 = 0
⇒ (2 sin x − 1)(sin x + 1)=0
⇒ sin x = −1, 1/2
⇒x = π/6,5π/6,3π/2 ∈ [0, 2π]
⇒ But for the case where x = 3π/2, tan x and sec x are not defined.
Ques. Determine the sign of cos 390°
- Positive
- Negative
- Both positive and negative
- Cannot be determined
Click here for the answer
Ans: (a) positive
Explanation: For the required angle, we first need to determine the coterminal angle which lies between 0° and 360°.
⇒ 400° < 390° < 40°
⇒ 390° lies in the fourth quadrant of the graph.
So, cos 410° is positive.
Ques: Determine the value of cos 765°.
- 1/√2
- 1
- 2
- √2
Click here for the answer
Ans: (a) 1/√2
Explanation: Since it is known that cosine function is periodic with period 2π
⇒ cos 765° = cos (720° + 45°)
⇒ cos (2 × 360° + 45°)
⇒ cos 45°
Therefore the result is 1/√2
Ques: A man observed a pole of height 30 ft. According to his measurement, the pole cast a 30 ft long shadow. Find the angle of elevation of the sun from the tip of the shadow using trigonometry.
- tan-1(1)
- -1
- tan-1(2)
- 0
Click here for the answer
Ans: (a) tan-1(1)
Explanation: Let x be the angle of elevation of the sun, then
⇒ tan x = 30/30 = 1
Therefore the result is x = tan-1(1)
Ques. The value of sin (60° + θ) – cos (60° – θ) is
- 2 cosθ
- 2 sinθ
- 1
- 0
Click here for the answer
Ans: (d) 0
Explanation: sin (60° + θ) – cos (60° – θ)
⇒ sin (60° + θ) – sin (90° -(60° – θ)) {since sin(90° – A) = cos A}
⇒ sin (60° + θ) – sin (60° + θ)
Therefore the result is 0
Ques: If the value of sec θ + tan θ = 81, then evaluate sec θ – tan θ.
- 1/81
- 6
- 7
- 10/81
Click here for the answer
Ans: (a) 1/64
Explanation: We know, sec2θ – tan2θ =1
⇒ (sec θ + tan θ)( sec θ – tan θ)=1
⇒ 81(sec θ – tan θ)=1
⇒ Or, sec θ – tan θ =1/81
So, the value of sec θ – tan θ =1/81
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:
Comments