Test: Trigonometric Identities - 1 - Class 10 MCQ

# Test: Trigonometric Identities - 1 - Class 10 MCQ

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## 15 Questions MCQ Test The Complete SAT Course - Test: Trigonometric Identities - 1

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Test: Trigonometric Identities - 1 - Question 1

### sec x + tan x = 2, find the value of cos x

Detailed Solution for Test: Trigonometric Identities - 1 - Question 1

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 = 1/2    ....(ii)
Adding the equation (i) and (ii)
2 sec x = 2 + 1/2
2/cos⁡x = 5/2
cos x = 4/5

Test: Trigonometric Identities - 1 - Question 2

### If Tan4θ + Tan2θ = 1, then what is the value of Cos4θ + Cos2θ?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 2

Given:
Tan4θ + Tan2θ = 1
Formula:
Sec2θ - Tan2θ = 1
Sin2θ + Cos2θ = 1
Calculation:
Tan4θ + Tan2θ = 1
⇒ Tan2θ (Tan2θ + 1) = 1
⇒ Tan2θ. Sec2θ = 1      {∵ sec2θ = 1 + tan2θ}
⇒ (sin2 θ/cos4 θ) = 1
⇒ 1 – cos2θ = cos4 θ      {∵ sin2 θ = 1 – cos2θ}
⇒ cos4 θ + cos2 θ = 1
∴ Then the value of cos4 θ + cos2 θ is 1.

Test: Trigonometric Identities - 1 - Question 3

### If tan 48° tan 23° tan 42° tan 67° = tan(A + 30°) then A will be

Detailed Solution for Test: Trigonometric Identities - 1 - Question 3

Formula used:
tan(90° - θ) = cot θ
tan θ × cot θ = 1

Calculation:
Given that,
tan 48° tan 23° tan 42° tan 67° = tan(A + 30∘)
⇒ tan (90° - 42°)tan(90 - 67°)tan 42°tan 67° = tan(A +  30°)
⇒ cot 42° cot 67° tan 42°tan 67° = tan(A + 30°)
∵ tan θ × cot θ = 1
⇒ 1 × 1 = tan(A + 30°)
⇒ tan 45° = tan(A + 30°)    (∵ tan 45° = 1)
⇒ 45° = A + 30°
⇒ A = 15°

Test: Trigonometric Identities - 1 - Question 4

sec4 x - tan4 x is equal to ?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 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

Test: Trigonometric Identities - 1 - Question 5

If sin θ + cos θ = 7/5, then sinθ cosθ is?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 5

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
∴ sin θcos θ = 12/25

Test: Trigonometric Identities - 1 - Question 6

If sin θ + cos θ = m and sec θ + cosec θ = n, then find the value of n(m + 1)(m - 1).

Detailed Solution for Test: Trigonometric Identities - 1 - Question 6

Calculation:
m = sin θ + cos θ
⇒ m2 = (sin θ + cos θ)2
⇒ m2 = sin2θ + cos2θ + 2sinθ.cosθ
⇒ m2 = 1 + 2sinθcosθ

⇒ n(m2 - 1) = 2m
⇒ n(m + 1)(m - 1) = 2m

Test: Trigonometric Identities - 1 - Question 7

If tan2θ = cot (3θ + 10°), then the value of θ equals:

Detailed Solution for Test: Trigonometric Identities - 1 - Question 7

Concept:
The question employs the concept of complementary trigonometric identities.
tanθ = cot (90 - θ)
sinθ = cos (90 - θ)
secθ = cosec (90 - θ)

Calculation:
Since tanθ and cot (90 - θ), forms a complementary pair-
⇒ 2θ + 3θ + 10°  = 90°
⇒ 5θ + 10° = 90°
⇒ 5θ = 80°
∴ θ = 16°
The value of θ is 16°

Test: Trigonometric Identities - 1 - Question 8

If 2sinθ/cos3θ = tan 270° - tan θ, find the value of θ?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 8

Formula:
sin2θ = 2sinθcosθ
sin(A - B) = sinAcosB - cosAsinB

Calculation:
Multiplying the left hand side with cosθ, we get-

sin2θ can also be written as sin(3θ - θ)

Applying sin(A-B) formula in the numerator part, we get-

⇒ tan3θ - tanθ
Equating LHS with RHS, we get-
⇒ tan3θ - tanθ = tan 270° - tan θ
⇒ 3θ = 270°
⇒ θ = 90°
∴ The value of θ is 90°

Test: Trigonometric Identities - 1 - Question 9

What is equal to?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 9

Formula Used:

a2 - b2 = (a - b) (a + b)
Calculation:
We have to find the value of

tan θ & cot θ can be written as,

Hence,
(sin θ + cos θ)

Test: Trigonometric Identities - 1 - Question 10

What is the value of sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360°?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 10

Formula used:
sin(360° - θ) = - sin θ

Calculation:
sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360°
⇒ sin 0° + sin 10° + ....+ sin 180° + sin (360 - 170°) +  sin (360 - 160°) + .....+ sin (360 - 10°) + sin (360 - 0°)

By using the  above formula
⇒ sin 0° + sin 10° + ......- sin 10° - sin 0° = 0
∴ sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360° = 0

Test: Trigonometric Identities - 1 - Question 11

Evaluate:

sin2 5° + sin2 10° + sin2 15° + …… + sin2 85° + sin2 90°

Detailed Solution for Test: Trigonometric Identities - 1 - Question 11

Concept:
I.
sin (90° - θ) = cos θ
II. sin2 θ + cos2 θ = 1

Calculation:
⇒ sin2 5° + sin2 10° + sin2 15° + …… + sin2 85° + sin2 90°
= (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + ….. +(sin2 40° + sin2 50°)+ sin2 45° + sin2 90°
= (sin2 5° + sin2 (90° - 5°)) + (sin2 10° + sin2 (90° - 10°)) + ….. +(sin2 40° + sin2 (90° - 40°))+ sin2 45° + sin2 90°
As we know that, sin (90° - θ) = cos θ
= (sin2 5° + cos2 5°) + (sin2 10° + cos2 10°) + ….. +(sin2 40° + cos2 40°)+ sin2 45° + sin2 90°
As we know that, sin2 θ + cos2 θ = 1
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1/√2)2 + 1
= 9 + (1/2) = 19/2

Test: Trigonometric Identities - 1 - Question 12

If tan θ + sec θ = 4, then find the value of cos θ ?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 12

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

Test: Trigonometric Identities - 1 - Question 13

If p = cosec θ – cot θ and q = (cosec θ + cot θ)-1 then which one of the following is correct?

Detailed Solution for Test: Trigonometric Identities - 1 - Question 13

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
1/q × p = 1
⇒ p = q

Test: Trigonometric Identities - 1 - Question 14

If tanA + tan(45) + tanC = tan(45) × tanA × tanC, then find the tan(A + C).

Detailed Solution for Test: Trigonometric Identities - 1 - Question 14

Given:
tanA + tan(45) + tanC = tan(45) × tanA × tanC,

Concept used:
(1.) If A + B + C = 180∘, tanA + tan(B) + tanC = tanA × tanB × tanC,
(2.) Sum of all internal angles of a triangle is 180∘.

Calculation:
According to the question,
⇒ tanA + tan(45) + tanC = tan(45) × tanA × tanC
As per the above concept.
⇒ A + B + C = 180
⇒ A + 45∘ + C = 180
⇒ A + C = 180- 45
⇒ A + C = 135
Therefore,
⇒ tan(A + C) = tan(135)
⇒ tan(135∘) = tan(90 + 45) = -1
Therefore, '-1' is the required answer.

Test: Trigonometric Identities - 1 - Question 15

If 3 - 4cotθ = cosecθ and 4 + 3cotθ = kcosecθ, find tha value of k

Detailed Solution for Test: Trigonometric Identities - 1 - Question 15

Formula:
1 + cot2θ = cosec2θ

Calculation:
Squaring both equations, we get-
⇒ (3 - 4cot)2 = (cosecθ)2
⇒ 9 + 16cot2θ - 2 × 3 × 4cotθ = cosec2θ
⇒ 9 + 16cot2θ - 24cotθ = cosec2θ (Eq 1)
Similarly,
⇒ (4 + 3cotθ)2 = (kcosecθ)2
⇒ 16 + 9cot2θ + 2 × 4 × 3cotθ = k2cosec2θ
⇒ 16 + 9cot2θ + 24cotθ = k2cosec2θ (Eq 2)
Adding Eq 1 and Eq 2, we get-
⇒  9 + 16cot2θ - 24cotθ + 16 + 9cot2θ + 24cotθ = cosec2θ + k2cosec2θ
⇒ 25 + 25cot2θ = cosec2θ (1 + k2)
⇒ 25 (1 + cot2θ) = cosec2θ (1 + k2)
Substituting the value of 1 + cot2θ as cosec2θ, we get-
⇒ 25 × cosec2θ = cosec2θ (1 + k2)
⇒ 25 = 1 + k2
∴ k2 = 24 or k = 2√6
The value of k is 2√6

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