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Introduction to Trigonometry Test - 5

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Introduction to Trigonometry Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

What will be the value of if = 1?

A
5

B
9

C
11

D
12

Solution

= 1 (Given)

cot = 1

Therefore, = 45°

Now,

= 3cot²45° - tan²45°+ 4sin²45°+ 5

= 3(1)²- (1)²+ 4+ 5
= 3 - 1 + 4 + 5

= 2 + 2 + 5 = 9
Question 2
1 / -0

(2 + ) - (2 + sin²) = __________.

A
sin²

B
cosec² - sin²

C
cosec²

D
cosec² + sin²

Solution

(2 + ) - (2 + sin²)

=

=

=

= cosec² - sin²
Question 3
1 / -0

In a right triangle PQR, right-angled at Q, sin P = and cos P = . What is the value of sec P cosec P?

A

B

C
5

D
2

Solution

sin P =
cos P =
We know, sec P =
sec P =
=
Similarly, cosec P =
=
=
sec P . cosec P =
=
Question 4
1 / -0

Simplify:

A
cosec

B
cot

C
-cot

D
sin

Solution

=

=

=

=

=
Question 5
1 / -0

Find the value of cosθ + if sec²θ - 1 = (1 + - )².

A
1

B

C
2

D

Solution

sec²θ - 1 = (1 + - )²
Assec²θ- 1 = tan²θ;
tan²θ = (1 + - )²

tan²θ = (1 + - )²
tan²θ = (1 + - (1 + ))²
tan²θ = (1 + - 1 - ))²
tan²θ = (0)²
tanθ = 0
Therefore, θ = 0°
Now,

= cos0°+

= 1 +

= 1 + 1 = 2
Question 6
1 / -0

What is the value of cos 0° . sin 25° . sec 85° . tan 45° . cosec 5° . cos 65° . sin 90°?

A
0

B
sin²25° . sec²85°

C
1

D
sin²25° . cos²85°

Solution

cos 0° . sin 25° . sec 85° . tan 45° . cosec 5° . cos 65° . sin 90°
= (1) sin 25° . sec 85° . (1) . sec(90° - 5°) . sin(90° - 65°) . (1)
= sin 25° . sec 85° . sec 85° . sin 25°
= sin²25° . sec²85°
Question 7
1 / -0

If and , then z is equal to

A

B

C

D

Solution

... (1)
... (2)

Using (2) in (1),
Question 8
1 / -0

If 2tany – tan²y = 1, then equals

A
1

B
0

C
-1

D
-3

Solution

2 tan y – tan²y = 1 ... (i)
2 tan y = 1 + tan²y
2 tan y = sec²y ... (ii)

Using (ii),
= 4 tan²y + tan⁴y – 4 tan³y
= (2 tan y – tan²y)²
Using (i),
= 1
Question 9
1 / -0

What is the value of (A + B) if tan(C + 2B) = and sin(2A - C) =^_,here each of A, B and C measures less than 90°?

A
30°

B
45°

C
60°

D
90°

Solution

tan(C + 2B) =
tan(C + 2B) = tan 60°
C + 2B = 60° ... (1)
sin(2A - C) =
sin(2A - C) = sin 30°
2A - C = 30° ... (2)

Now, adding equations (1) and (2):
2A + 2B = 60°+ 30°
2(A + B) = 90°
(A + B) = 45°
Question 10
1 / -0

If sinθ - cotθ = 1 and cosec²θ = 1, then cosθ is equal to

A
1

B
0

C
-1

D
-1/2

Solution

Given: sinθ - cotθ = 1 .......(i)
cosec²θ = 1.......(ii)
∵ cosec²θ - cot²θ = 1
1 - cot²θ = 1 (From equation (ii))
cot²θ = 0
cotθ = 0
Put the value of cotθ in equation (i), we get
sinθ - 0 = 1
sinθ = 1
∵ sin²θ + cos²θ = 1
1 + cos²θ = 1
cos²θ = 0
cosθ = 0
Question 11
1 / -0

Find the value of.

A

B

C

D

Solution

=
= 2sec60° + 1 - (3tan30°)³
=
=
Question 12
1 / -0

Evaluate if cos = 0

A
-2

B
0

C
1

D
-1

Solution

=

Given: cos = 0
Question 13
1 / -0

What is the value of q²- p²- r²if p = acosecθ cotθ', q = acosecθ cosecθ' and r = acosecθ?

A
0

B
a²cosec²θ

C
cosec²θ

D
1

Solution

q²- p²- r²
= a²cosec²θcosec²θ' - a²cosec²θcot²θ' - a²cosec²θ
= a²cosec²θ(cosec²θ' - cot²θ') - a²cosec²θ
(As cosec²θ - cot²θ = 1)
= a²cosec²θ - a²cosec²θ = 0
Question 14
1 / -0

If = 3, = 2, then

A

B

C

D

Solution

... (1)

... (2)

Squaring and subtracting (2) from (1),
Question 15
1 / -0

What is the value of 'g' in cot²θ - cos²θ - g = (tanθ - cotθ)² - (secθ - cosθ)²?

A
0

B
1

C
-1

D
3

Solution

cot² - cos² - g = (tan - cot)² - (sec - cos)²
cot² - cos² - g = (tan² + cot² - 2tan cot) - (sec² + cos² - 2sec cos)
cot² - cos² - g = tan² + cot² - 2tan cot - sec² - cos² + 2seccos
cot² - cos² - g = -1 + cot² - cos² - 2 + 2
cot² - cos² - g = -1 + cot² - cos²
Therefore, comparing both sides, g = 1
Question 16
1 / -0

If tanA : secA = 11 : 9, then the value of is

A

B

C

D

Solution

=

=

=

=
Question 17
1 / -0

What is the value of

A
Not defined

B
2

C
1

D
0

Solution

=
=
= 1 ÷ 1 - 1
= 1 - 1
= 0
Question 18
1 / -0

If X + Y = 90°, thenis equal to

A

B

C

D

Solution

=

=

=
=
=
Question 19
1 / -0

If sin + cos = a and sin – cos = b, then 4a²b² + (b² - a²)²equals

A
4

B
6

C
8

D
10

Solution

=
=
=
=
=
= 4
Question 20
1 / -0

In a triangle PQR, ∠Q = 90° and cosecR =. Find the value of .

A
8

B
12

C
16

D
20

Solution

Consider the triangle PQR in which ∠Q = 90^o and cosecR =^.

Let PQ = √3x, PR = 2x

By Pythagoras theorem,

RQ = x

=
== 16
Question 21
1 / -0

What is the value of (x²- y²) - 4sec tan if x = sec + tan and y = sec - tan?

A
1

B
8 sec tan

C
-1

D
0

Solution

(x²- y²)- 4sec tan
= [(sec + tan)²- (sec - tan)²] - 4sec tan
= [(sec² + tan² + 2sec tan) - (sec² + tan² - 2sec tan )] - 4sec tan
= [sec² + tan² + 2 sec tan - sec² - tan² + 2sec tan] - 4sec tan
= 4sec tan - 4sec tan
= 0
Question 22
1 / -0

Choose the INCORRECT statement(s).

I. The expression is dependent on .

II. If sin + cos = 0, then .

A
Only I

B
Only II

C
Neither I nor II

D
Both I and II

Solution

I.

=

=

=

= 3

That means the given expression is independent of Θ.

Hence, statement I is incorrect.

II.
LHS
=

=

=

Therefore , LHS = RHS

Hence, statement II is correct.
Question 23
1 / -0

Fill in the blanks:

A. = __M__

B. tan² - = _____N______

C. If l = x²sin³ and m = y² cos³, then the value of is _____O_____.

A

M N O

cos2 0 1

B

M N O

-1 1

C

M N O

cos2 1 -1

D

M N O

sin -1 1

Solution

A.

=

= [As 1 - sin² = cos²]

=

B. tan² -

= - [As tan = ]

= = [As sin² + cos² = 1]

= -1

C.

Now, putting the values of l and m,

= sin⁴ + cos⁴ + 2sin² cos²

= (sin²)² + (cos²)² + 2sin² cos²

= (sin² + cos²)²

[As a²+ b²= (a + b)²- 2ab)]

= 1 [As sin² + cos² = 1]
Question 24
1 / -0

If , and , then the value of is

A

B

C

D

Solution

Square both sides, we get

Square both sides, we get
So,
Question 25
1 / -0

Which of the following statements is/are true?

(a) = 5

(b) If A + B = 90°, then.

A
Only (a)

B
Only (b)

C
Both (a) and (b)

D
Neither (a) nor (b)

Solution

(a) LHS
=

=

= 1 + 1 + 1 = 3

Therefore, LHS is not equal to RHS.

(b)
RHS =
=

=

=

Therefore, LHS is not equal to RHS.

Hence, neither of the statements is true.

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