Test: Basic Trigonometric Formula - Commerce MCQ

# Test: Basic Trigonometric Formula - Commerce MCQ

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## 15 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Basic Trigonometric Formula

Test: Basic Trigonometric Formula for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Basic Trigonometric Formula questions and answers have been prepared according to the Commerce exam syllabus.The Test: Basic Trigonometric Formula MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Trigonometric Formula below.
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Test: Basic Trigonometric Formula - Question 1

### cos 68° cos 8° + sin 68° sin 8° = ?

Detailed Solution for Test: Basic Trigonometric Formula - Question 1

We know,
cosA cosB + sinA sinB = cos(A-B)

cos 68° cos 8° + sin 68° sin 8° = Cos (68-8) = Cos60°
=1/2

Test: Basic Trigonometric Formula - Question 2

### In a triangle ABC, tan A/2 = 5/6, tan B/2 = 20/37, then tan C/2 is equal to:

Detailed Solution for Test: Basic Trigonometric Formula - Question 2

In triangle ABC,
► Sum of all three angles = 1800
► A + B + C = 180

Test: Basic Trigonometric Formula - Question 3

### In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals:

Detailed Solution for Test: Basic Trigonometric Formula - Question 3

cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

Test: Basic Trigonometric Formula - Question 4

The value of Cos75°is equal to

Detailed Solution for Test: Basic Trigonometric Formula - Question 4

Test: Basic Trigonometric Formula - Question 5

Sin A = 1/√10 , Sin B = 1/√5 If A and B are both acute angles, then, A+B =?

Detailed Solution for Test: Basic Trigonometric Formula - Question 5

We know that:
Sin θ = Opposite / Hypotenuse

∴ SinA = 1/√10
CosA= 3/√10
similarly, SinB = 1/√5
CosB= 2/√5

Multiply:
Cos(A+B)= CosA x CosB - SinA x SinB

Substituting the value in above equation we get:

= 3/√10 x 2/√5 - 1/√10 x 1/√5
= 6/√50 - 1/√50
= 6-1/5√2. ........(√50=5√2)
= 1/ √2

we know that, sin 45 = 1/ √ 2 therefore
sinθ / cosθ = 45

Test: Basic Trigonometric Formula - Question 6

If 3 × tan(x – 15) = tan(x + 15), then the value of x is:

Detailed Solution for Test: Basic Trigonometric Formula - Question 6

3 × tan (x – 15) = tan (x + 15)
⇒ tan(x + 15) / tan(x – 15) = 3/1

► {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = (3 + 1) / (3 – 1)

► {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = 2

sin(x + 15 + x – 15) / sin(x + 15 – x + 15) = 2
sin 2x / sin 30 = 2
sin 2x / (1/2) = 2
2 × sin 2x = 2
Sin 2x = 1

∴ Sin 2x = Sin 90
2x = 90
x = 45

Test: Basic Trigonometric Formula - Question 7

The value of tan 20 × tan 40 × tan 80 is

Detailed Solution for Test: Basic Trigonometric Formula - Question 7

Test: Basic Trigonometric Formula - Question 8

Chose which of the following expressions equals sinA + cosA.

Detailed Solution for Test: Basic Trigonometric Formula - Question 8

sinA + cosA = √2(sinA/√2+cosA/√2)
=√2(sinAcos(π/4)+cosAsin(π/4))
=√2sin(A+π/4)
by using sin(A+B) formula

Test: Basic Trigonometric Formula - Question 9

If acos x + bsin x = c, then the value of (asin x – bcos x)²  is:

Detailed Solution for Test: Basic Trigonometric Formula - Question 9

(acos x + bsin x)² + (asin x – bcos x)² = a² + b²
⇒ c² + (asin x – bcos x)² = a² + b²
⇒ (asin x – bcos x)² = a² + b² – c²

Test: Basic Trigonometric Formula - Question 10

sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to:

Detailed Solution for Test: Basic Trigonometric Formula - Question 10

L.H.S. = sin(60+A)cos(30−B)+cos(60+A)sin(30−B)
= sin[(60+A)+(30−B)]            (Using, sin(A+B)sinAcosB+cosAsinB)
= sin(90+A−B)
= sin(90+(A−B))
= cos(A−B)            (Using, sin(90+θ)=cosθ)       = R.H.S.Hence Proved.

Test: Basic Trigonometric Formula - Question 11

If cos a + 2cos b + cos c = 2 then a, b, c are in

Detailed Solution for Test: Basic Trigonometric Formula - Question 11

cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)

2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)

2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)

2 sin(A/2) sin(C/2) = sin(B/2)
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac

2(s – b) = b
a + b + c – 2b = b
a + c – b = b
a + c = 2b

Test: Basic Trigonometric Formula - Question 12

sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A =

Detailed Solution for Test: Basic Trigonometric Formula - Question 12

sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A

Test: Basic Trigonometric Formula - Question 13

cos(π/4 - x) cos (π/4 - y) -sin(π/4 - x) sin(π/4 - y) =

Detailed Solution for Test: Basic Trigonometric Formula - Question 13

Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)

= CosA*Cos B - Sin A*Sin B
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)

Test: Basic Trigonometric Formula - Question 14

The value of tan 3A – tan 2A – tan A is

Detailed Solution for Test: Basic Trigonometric Formula - Question 14

3A= A+ 2A
⇒ tan 3A = tan (A + 2A)
⇒ tan 3A = (tan A + tan 2A) / (1 – tan A . tan 2A)
⇒ tan A + tan 2A = tan 3A – tan 3A x tan 2A . tan A
⇒ tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Test: Basic Trigonometric Formula - Question 15

Which of the following is incorrect:

Detailed Solution for Test: Basic Trigonometric Formula - Question 15

D is not a trigonometric identity.

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85 videos|243 docs|99 tests

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