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Introduction to Trigonometry Test - 6

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Introduction to Trigonometry Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of .

A

B

C

D

Solution
Question 2
1 / -0

If cosθ = 2cos²θ , then 2sec²θ – cos²θ + tan²θ =

A

B

C

D

Solution

cosθ = 2 cos²θ

secθ = 2
secθ = sec 60°
θ = 60°
2 sec²θ – cos²θ + tan²θ= 2 sec²60° - cos²60° + tan²60° =
=
=
Question 3
1 / -0

is equal to

A
secθ.cosθ

B
secθ.cotθ

C

D
sec²θ - cot²θ

Solution

=

=

=

= secθ.cotθ
Question 4
1 / -0

If xcotθ - ycosecθ = a and xcosecθ - ycotθ = b, then (y² - x²)² is equal to

A
2(a² - b²)²

B
(a - b)²

C
(a² + b²)²

D
(a² - b²)²

Solution

xcotθ – ycosecθ = a
Squaring both sides, we get
x²cot²θ + y²cosec²θ – 2.xy.cotθ.cosecθ = a² --- (1)
xcosecθ – ycotθ = b
Squaring both sides, we get
x²cosec²θ + y²cot²θ – 2.xy.cosecθ.cotθ = b² --- (2)
Subtracting (2) from (1), we get
x²cot²θ + y²cosec²θ – 2.xy.cotθ.cosecθ - (x²cosec²θ + y²cot²θ – 2.xy.cosecθ.cotθ) = a² - b²
x²cot²θ + y²cosec²θ - x²cosec²θ - y²cot²θ = a² - b²
cosec²θ(y² – x²) – cot²θ(y² – x²) = a²– b²
(y² – x²)(cosec²θ – cot²θ) = a² – b²
(y² – x²) = a² – b²
Squaring both sides,
(y² – x²)² = (a² - b²)²
Question 5
1 / -0

Simplify

A

B

C

D

Solution

=

=
=

=

=
Question 6
1 / -0

If tan A + cot A = 4, then what is the value of tan³ A + cot³ A?

A
52

B
64

C
48

D
72

Solution

According to the question,
(a + b)³ = a³ + b³ + 3ab (a + b)
∴ a³ + b³ = (a + b)³ - 3ab (a + b)
So, tan³ A + cot³ A = (tan A + cot A)³- 3 . tan A . cot A (tan A + cot A)
= (4)³ - 3 × 4= 64 - 12 = 52
Question 7
1 / -0

If , then equals

A

B

C

D

Solution

=

=

=

=

=

=

=
Question 8
1 / -0

If x + y = 90°, then what is the value of ?

A
cos x

B
sin x

C

D

Solution

( x + y = 90°, Given)

=

=

=
Question 9
1 / -0

If A + B = 90°, sin A = and cos B = , find the relation between p, q, r and s.

A
ps = qr

B
p² + s² = q² + r²

C
pq = rs

D
pr = qs

Solution

A + B = 90°
B = 90° - A
sin A = ……..(1)
cos B =
cos (90° - A) =
sin A = ……..(2)
From (1) and (2)
= ps = rq
Option (1) is correct.
Question 10
1 / -0

In triangle RPS, it is given that ∠P = 90° and cosecR = . Find the value of [tanS.cotR – secR.sinR + cosR]³.

A

B

C

D

Solution

cosecR = .

cosec R = cosec 45°
R = 45°
P = 90°
So, S = 45°
[tanS.cotR – secR.sinR + cosR]³= (tan 45° cot 45° - sec45° sin45° + cos45°)³
=
=
=
Question 11
1 / -0

If sin x + sin² x = 1, then find the value of cos⁴ x + cos⁸ x + 2 cos⁶ x + 1.

A
-1

B
0

C
1

D
2

Solution

sin x = 1 - sin² x = cos² x
cos⁴ x + cos⁸ x + 2cos⁶x + 1
= sin² x + sin⁴x + 2 sin³ x + 1
= (sin x + sin²x)² + 1
= 2
Question 12
1 / -0

Evaluate: 21 cot48° cot81° cot42° cot9° - 9()³+ 16()²

A
28

B
27

C
26

D
22

Solution

21 cot48° cot81° cot42° cot9° - 9()³+ 16()²

= 21 cot(90° - 42°) cot(90° - 9°) cot42° cot9° - 9()³ + 16()²= 21 tan42° tan9° cot42° cot9° - 9()³+ 16()²[As cot(90° - θ) = tanθ, sec(90° - θ) = cosecθ, sin(90° - θ) = cosθ]

= 21cot42° cot9° - 9(1)³ + 16(1)²= 21 – 9 + 16 = 12 + 16 = 28
Question 13
1 / -0

What is the value of q if cot(p + q – r + s) = 0, sin(p – q) = , cosec(q + s) = and sec(q + r) = 1?

A
9.5°

B
8.7°

C
7.5°

D
6°

Solution

cot(p + q – r + s) = 0
cot(p + q – r + s) = cot 90°
Therefore, (p + q – r + s) = 90° --- (1)
Now, sin(p – q) =

sin(p – q) = sin 30°
So,(p – q) = 30° --- (2)
Also, cosec(q + s) =
cosec(q + s) = cosec 45°
(q + s) = 45° --- (3)
And sec(q + r) = 1
sec(q + r) = sec 0°
(q + r) = 0° --- (4)
Now, adding equations (2) and (4),
p + r = 30° --- (5)
Now, putting (3) in (1),
So, p – r = 90° - 45° = 45° --- (6)
Now, subtracting equation (6) from equation (5),
2r = -15°
r = -7.5°
Now, putting r = -7.5° in equation (4), we get
q = 7.5°
Question 14
1 / -0

If cosθ =sinθ, then the value of+is

A

B

C

D

Solution

Gievn: cosθ =sinθ
= --- (1)

Now, +

Dividing both numerator and denominator by sinθ,

= +

= + [From (1)]

= +

= +

=

= =
Question 15
1 / -0

What is the value of p in the following equation?

p – sec⁴θ – cosec⁴θ = 2(tanθ + cotθ)² - (sec²θ + cosec²θ)²

A
3

B
2

C
1

D
0

Solution

R.H.S.
2(tanθ + cotθ)²- (sec²θ + cosec²θ)²
= 2- (sec⁴θ + cosec⁴θ + 2 sec²θcosec²θ)

= 2()²- (sec⁴θ + cosec⁴θ + 2 sec²θcosec²θ)

= 2sec²θcosec²θ - sec⁴θ - cosec⁴θ - 2sec²θcosec²θ

= -sec⁴θ – cosec⁴θ

Comparing this with L.H.S, we get

p = 0
Question 16
1 / -0

What is the value of ?

A
3/2

B
2

C

D
1/2

Solution

Therefore, option (1) is correct.
Question 17
1 / -0

What is the value of ?

A

B

C

D

Solution

Given:

=

=

=

=

=

=
Question 18
1 / -0

=

A

B

C

D

Solution

=

=

= [As ]

=
Question 19
1 / -0

If sin A + cosec A = 3, then find the value of .

A
1

B
0

C
7

D
3

Solution

sin A + cosec A = 3
sin A + = 3

Squaring both sides:
sin² A + + 2 = 9
= 9 - 2 = 7
Question 20
1 / -0

What is the value of if ?

A

B

C

D

Solution

Squaring both sides,

[As ]
Question 21
1 / -0

Which of the following statements is/are false?

(A) If secθ + cosecθ = 1, then .

(B).

A
Only (A)

B
Only (B)

C
Neither (A) nor (B)

D
Both (A) and (B)

Solution

(A) secθ + cosecθ = 1 --- (1)

L.H.S. =

=

=

= cotθ + secθ.tanθ + tanθ.cosecθ
= cotθ + tanθ(secθ + cosecθ)
= cotθ + tanθ [Using 1]

=

=

Multiply numerator and denominator by 2

=

=

=

Therefore, L.H.S. is not equal to R.H.S.
Hence, statement (A) is false.

(B) R.H.S. =

=

=

= sinR + secR
Therefore, L.H.S. is not equal to R.H.S.
Hence, statement (B) is false.
Question 22
1 / -0

Which of the following statements is/are CORRECT?

(I) ^{is dependent on θ.}
(II) .

A
Only I

B
Only II

C
Neither I nor II

D
Both I and II

Solution

Hence, statement I is correct as it is dependent on θ.

(II) R.H.S. =

=

=

=

=

= sin²θ – cosθ

Therefore, L.H.S. is not equal to R.H.S.
Hence, statement II is incorrect.
Question 23
1 / -0

If 4 cos² A + 2 sin² A = 3, then what is the value of A, given that A lies in the first quadrant?

A
45°

B
60°

C
30°

D
15°

Solution

4 cos² A + 2 (1 - cos² A) _{^{= 3}}

4 cos² A + 2 - 2cos²A = 3

2 + 2 cos²A = 3
2(1 + cos² A)= 3
1 + cos² A =
cos² A = – 1
=
=

cosA =
A = 45°
Question 24
1 / -0

What is the value of x²- 4 - y²+if x = secθ – cosecθ and y = tanθ – cotθ?

A

B
cotθ

C
sinθ - 2

D

Solution

x² – 4 - y²+

After putting the values of x and y, the expression becomes:

(secθ – cosecθ)² – 4 – (tanθ – cotθ)²+
= (sec²θ + cosec²θ - 2secθ cosecθ) - 4 - (tan²θ + cot²θ - 2tanθ cot θ) +

= sec²θ + cosec²θ - 2secθ cosecθ – 4 - tan²θ - cot²θ + 2tanθ cot θ +
= 1 + 1 -- 4 + 2 cotθ +[As sec²θ - tan²θ = 1 and cosec²θ - cot²θ = 1]

Now, 2 -- 2 +

=
Question 25
1 / -0

Fill in the blanks:

(i) If sec⁴θ + sec²θ = 1, then the value of 3 tan²θ + tan⁴θ is ____________.
(ii) The value of+, if p = m tan⁶θ and q = n cot⁶θ, is ___________.
(iii) If o = g cosecθ, m = gcotθ and n = Isecθ, then the value of o²– m²– n²+ I²tan²θ is ________.

A
(i) 1, (ii) , (iii) g²- I²

B
(i) 0, (ii) cos²θ, (iii) I

C
(i) 2, (ii), (iii) g²

D
(i) -1, (ii) ^,(iii) g²- I²

Solution

(i) sec⁴θ + sec²θ = 1 [Given]
sec²θ = 1 - sec⁴θ
1 + tan²θ = 1 – (1 + tan²θ)²[As sec²θ = 1 + tan²θ]
1 + tan²θ = 1 – [1 + tan⁴θ + 2tan²θ]
1 + tan²θ = 1 – 1 - tan⁴θ – 2tan²θ
1 + 3tan²θ + tan⁴θ = 0
3tan²θ + tan⁴θ = -1

(ii) +
Putting the values of p and q in this expression,
+

= tanθ + cotθ

= +

=

= [As sin²θ + cos²θ = 1]

(iii) o²– m²– n²+ I²tan²θ
Putting the values of o, m and n in this expression,
= g²cosec²θ - g²cot²θ - I²sec²θ + I²tan²θ
= g²(cosec²θ - cot²θ) - I²(sec²θ - tan²θ)
= g²- I²[As cosec²θ - cot²θ = 1 and sec²θ - tan²θ = 1]

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