CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 02

Question. If cosα /cosβ = mand  cosα /sinβ =n,(m² + n²)cos2β =?   
A) 2 m
B) 2 n
C) (m + n)
D) (m - n)

Answer: B

Question. If 7cosecΦ -3cotΦ = 7, then 7cotΦ - 3cosecΦ?   
A) 1
B) 2
C) 3
D) 4

Answer: C

Question. The value of 2(sin6Φ + cos6Φ) -3(sin4Φ + cos4Φ) = ?   
A) 0
B) 1
C) 2
D) 3

Answer: A

Question. If tan = 5/6 θ+Φ = 90 the value of cotΦ ?   
A) 1/6
B) 3/6
C) 5/6
D) 7/6

Answer: C

Question. If A,B are acute angle and sinA=cosB, then the value of A+B=?   
A) 300
B) 600
C) 900
D) 700

Answer: C

Question. If tan5Φ =1, then Φ ?   
A) 80
B) 70
C) 60
D) 90

Answer: D

Question. If sinΦ/1+cosΦ + 1+cosΦ /sinΦ = 4, Then Φ = ?   
A) 10⁰
B) 20⁰
C) 30⁰
D) 0

Answer: C

Question. The value of tanΦ in terms of sinΦ =?   
A) sinΦ /√(1+sin²Φ )
B) sin2Φ /1-sinΦ
C) sin2Φ /1+sinΦ
D) sinΦ /√(1+sin²Φ )

Answer: D

Question. If secΦ + tanΦ = 4sinΦ &cosΦ = ?   
A) 2/17 , 3/17
B) 15/17 , 8/17
C) 5/6
D) 7/6

Answer: B

VERY SHORT ANSWER TYPE QUESTIONS

Question. If cosec θ = 15/7 and θ + α = 90° , find the value of sec θ.

Answer: 15 / 7

Question. If A, B and C are the angles of a triangle, then find the value of cot (B + C)/2 in terms of A.

Answer: tan A/2

Question. What is the value cos 1° cos 2° cos 3° .... cos 89° cos 90° ...... cos 180°?

Answer: 0

Question. If 7 sin² θ + 3 cos²θ = 4, find the value of tan θ if 0° ≤ θ ≤  90°.

Answer: 1/√3

Question. Find the value of cosθ - cos³θ/sinθ - sin³θ .

Answer: tan θ

Question. If tanA = 1/√3 and sinB = 1/√2 , find the value of A + B.

Answer: 75°

Question. Write the value of sin2 68° + sin² 22° – 1.

Answer: 0

Question. If cos 3θ = 1, then find the value of tan θ.

Answer: 0

Question. If A and B are acute angles and sin B = cos A, then write the value of A + B.

Answer:  90°

Question. Write the value of sin (55° + θ) – cos (35° – θ).

Answer: 0

Question. Express tan 87° + sin 63° in terms of trigonometric ratios of angles between 0° and 45°.

Answer: cot 3° + cos 27°

Question. Write the value of tan 5° tan 35° tan 45° tan 55° tan 85°.

Answer: 1

Question. If tan (θ – 36°) = cot 2θ; 2θ and (θ – 36°) are acute angles, then find θ.

Answer:  42°

Question. If tan² θ – 5 tan θ + 1 = 0, find the value of tanθ + cot θ.

Answer: 5

Question. Write the simplest form of sinθ/sec(90° - θ) + cosθ/cosec (90° - θ)

Answer: 1

Question. What is the maximum value of 1/cosec θ ?

Answer: 1

Question. If sin 2θ = √3/2 , find the value of cos θ .

Answer: √3/2

Question. If sin θ+ cosθ = 1, the find the value of sin θ cos θ.

Answer: 0

Question. If cosec²θ (1 + cos θ) (1 – cos θ) = k, then find the value of k.

Answer: 1

Question. If sin 2θ = cos 3θ, then find the value of θ.

Answer: 18°

Question. Evaluate each of the following :
(i) tan 5° tan 25° tan 30° tan 45° tan 65° tan 85° (ii) cot 12° cot 38° cot 52° cot 60° cot 78°
(iii) sec (35° + A) – cosec (55° – A) (iv) cos (36° + A) – sin (54° – A)
Solution. (i) 1/√3 (ii) 1/√3 (iii) 0 (iv) 0

Question. Evaluate each of the following :
(i) sin 54° – cos 36° (ii) tan 62° – cot 28°
(iii) cosec 47° – sec 43° (iv) sec² 31° – cot² 59°
(v) sin² 29° + sin² 61 (vi) tan² 48° – cosec²42°
Solution. (i) 0 (ii) 0 (iii) 0 (iv) 1 (v) 1 (vi) –1

Question. If 21 cosec θ = 29, find the value of : 
(i) cos² θ - sin ²θ/1 - 2sin² θ    (ii) 2cos² θ - 1/cos² θ - sin² θ 
Solution. (i) 1 (ii) 1

Question. If sin A - 1/sin A = 3 , find the value of sin2 A + 1/sin² A.
Solution. 11

Question. Given that sin (A + B) = sin A cos B + cos A sin B, find the value of sin 75°.
Solution. √3 + 1/2√2

Question. Given that cos (A – B) = cos A cos B + sin A sin B, find the value of cos 15°.
Solution. √3 + 1/2√2

Question. Find x in each of the following, if :
(i) 2 cos x = 1 (ii) 2 sin 2x = 1 (iii) tan 3x =√3 (iv) 2sin 3x =√3  (v) √3 cot 3x = 1  (vi) √3.sec(x/) = 2
Solution. (i) 60° (ii) 15° (iii) 20° (iv) 20° (v) 20° (vi) 60°

Question. Find acute angle θ in each of the following cases; if,
(i) sin (3θ – 15°) = 1 (ii) 2 sin (3θ – 15°) √3  (iii) sin2 θ = 1/4  (iv) 3tan2 θ - 1 = 0 (v) cos (40° + θ) = sin 30°  (vi) cot2 (2θ - 5°) = 3
Solution. (i) 35° (ii) 25° (iii) 30° (iv) 30° (v) 20° (vi) 17.5°

Question. If sin (A 2B) = √3/2 and cos (A + 4B) = 0, find the values of angles A and B.
Solution. A = 30°, B = 15°

Question. If sin (A + B) = 1 and cos (A B) = √3/2 ; 0° <A + B ≤ 90° , and A > B, find A and B.
Solution. A = 60°, B = 30°

Question. ABC is a right triangle, right angled at C. If A = 30°, and AB = 40 units, find the remaining two sides and ∠B of ΔABC. 
Solution. AC = 20 √3 units, BC = 20 units and B = 60°

Question. If sin (A – B) 1/2 and cos (A + B) = 1/2 ; 0° <A + B < 90° , and A > B, find A and B.
Solution. A = 45°, B = 15°

Question. If tan θ = a/b , find the value of cos θ + sin θ/cos θ - sin θ .
Solution. b + a/ b - a 

Question. In ΔABC, right angled at B, AC + BC = 25 cm and AB = 5 cm, find the value of sin2A + cos2A.
Solution. 1

Question. In ΔABC, right angled at B, if AB = 12 cm and BC = 5 cm, find (i) sin A and tan A (ii) sin C and cot C.
Solution. (i) 5/13 , 5/12 (ii) 12/13 , 5/12

Question. If sec 5 θ = cosec (θ – 36°), where 5θ is an acute angle, find the value of θ.
Solution. 21°

Question. Using trigonometric identities, write the following expressions as an integer :
(i) 5 cot² A – 5 cosec² A (ii) 4 tan2 θ – 4 sec² θ
(iii) 7 cos²θ + 7 sin²θ (iv) 3 sec²α – 3 tan²α
Solution. (i) –5 (ii) –4 (iii) 7 (iv) 3

Question. If 2 tan θ = 1, find the value of 3cos θ + 2sinθ/2cos θ - sin θ .
Solution. 8/3

Question. Express each of the following in terms of trigonometric ratios of angles between 0° and 45° :
(i) sin 79° + cosec 83° (ii) tan 58° + sec 46°
(iii) cosec 57° + cot 57° (iv) sin 82° + tan 82°
Solution. (i) cos 11° + sec 9° (ii) cot 32° + cosec 44° (iii) sec 33° + tan 33° (iv) cos 8° + cot 8°

Question. If sin 3θ = cos (θ – 6°), where 3θ and θ – 6°) are acute angles, find the value of θ.
Solution. 24°

Question. If cot θ =13/12 , find the value of 2sin θ cos θ/cos2 θ - sin² θ .
Solution. 312/25

Question. If tan²θ = cot (θ + 6°), where 2θ and θ + 6° are acute angles, find the value of θ.
Solution. 28°

Question. Prove that: 2 sec ² θ - sec ⁴ θ - 2 cosec ² θ + cosec ⁴ θ = cot ⁴ θ -tan⁴ θ.
Sol. LHS = 2 sec ² θ - sec ⁴ θ - 2 cosec ² θ + cosec ⁴ θ
             = 2 (sec ² θ) - (sec ² θ)2 - 2 (cosec ² θ) + (cosec 2 θ)2
             = 2 (1 + tan2 θ) - (1 + tan² θ)2 - 2 (1 + cot ² θ) + (1 + cot ² θ)2
             = 2 + 2 tan² θ - (1 + 2 tan² θ + tan⁴ θ) - 2 - 2 cot ² θ + (1 + 2 cot ² θ + cot ⁴ θ)
             = 2 + 2 tan² θ -1 - 2 tan² θ - tan⁴ θ - 2 - 2 cot ² θ +1 + 2 cot ² θ + cot 4 θ
             = cot ⁴ θ - tan⁴ θ = RHS.

Question. Prove that: (1 - sin A + cos A)² = 2 (1 -sin A)(1 + cos A).
Sol. LHS = (1 - sin A + cos A)²
             = 1 + sin² A + cos² A - 2 sin A + 2 cos A - 2 sin A cos A
             = 1 +1 - 2 sin A + 2 cos A - 2 sin A cos A [Q sin² A + cos² A = 1 ]
             = 2 (1 - sin A + cos A - sin A cos A) = 2 [(1 - sin A) + cos A (1 - sin A)]
             = 2 (1 - sin A) (1 + cos A) = RHS.

Question. If x sin³θ + y cos³θ = sinθ cosθ and xsinθ = ycos θ, prove x² + y² =1.
Sol. We have,
x sin³ θ + y cos³ θ = sin θ cos θ
⇒ ( sin x θ) sin² θ + (y cos θ) cos² θ = sin θ cos θ
⇒ x sin θ (sin² θ) + (x sin θ) cos² θ = sin θ cos θ          [∴ x sin θ = y cos θ]
⇒ x sin θ (sin² θ + cos² θ) = sin θ cos θ
⇒ x sin θ = sin θ cos θ ⇒ x = cos θ
Now, we have x sin θ = y cos θ
⇒ cos q sin θ = y cos θ                                                [∴ x = cos θ]
⇒              θ = sin θ
Hence, x ² + y² = cos² θ + sin² θ =1.

Question. Prove that: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan2 A + cot ² A. 
Sol. LHS = (sin A + cosec A)² + (cos A + sec A)²
= sin2 A + cosec ² A + 2 sin A . cosec A + cos² A + sec ² A + 2 cos A . sec A
= (sin² A + cosec ² A + 2) + (cos² A + sec ² A + 2)                                     {sinA . cosecA = 1}
                                                                                                                    {cosA . secA = 1}
= (sin² A + cos² A) + (cosec ²A + sec ² A) + 4
=1 +1 + cot ² A +1 + tan² A + 4
= 7 + tan² A + cot ² A = RHS.