Exercise 8.1

1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

Answer:

Let us draw a right angled triangle ABC, right angled at B.

Using Pythagoras theorem,

(i) 

(ii)

,

2. In adjoining figure, find tan P – cot R.

Answer:

3. If sin A =3/4, calculate cos A and tan A.

Answer:

Given: A triangle ABC in which

B =90

We know that sin A = BC/AC = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
AC² = AB² + BC²
(4k)² = AB² + (3k)²
16k² – 9k² = AB²
AB² = 7k²
AB = √7 k
cos A = AB/AC = √7 k/4k = √7/4
tan A = BC/AB = 3k/√7 k = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Answer:

Let ΔABC be a right-angled triangle, right-angled at B.
We know that cot A = AB/BC = 8/15   (Given)
Let AB be 8k and BC will be 15k where k is a positive real number.
By Pythagoras theorem we get,
AC² = AB² + BC²
AC² = (8k)² + (15k)²
AC² = 64k² + 225k²
AC² = 289k²
AC = 17 k
sin A = BC/AC = 15k/17k = 15/17
sec A = AC/AB = 17k/8 k = 17/8

5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Answer:

Consider a triangle ABC in which

Let AB = 12k and AC = 13k

Then, using Pythagoras theorem,

(AC)2 = (AB)2 + (BC)2

(13 k)2 = (12 k)2 + (BC)2

169 k2 = 144 k2 + BC2

25 k2 = BC2

BC = 5 k

 (I)   Sin  = 

=  = 

(ii) Cos θ = 

=   = 

(iii)Tan θ = 

=  = 

(iv) Cot θ = 

=  = 

(v) Cosec θ = 

=  = 

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Answer:

cosA=cos

But 

7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot²θ

Answer:

Consider a triangle ABC

 

 (ii)

 

8. If 3cot A = 4/3 , check whether (1-tan²A)/(1+tan²A) = cos²A – sin²A or not.

Answer:

Consider a triangle ABC AB=4cm, BC= 3cm

.

And 

9. In triangle ABC, right-angled at B, if tan A =1/√3 find the value of:

(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Answer:

Consider a triangle ABC in which

.

(i)

(ii)

10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer:

Given that, PR + QR = 25 , PQ = 5
Let PR be x.  ∴ QR = 25 – x
By Pythagoras theorem ,
PR2 = PQ² + QR²
x² = (5)2 + (25 – x)²
x² = 25 + 625 + x² – 50x
50x = 650
x = 13
∴ PR = 13 cm
QR = (25 – 13) cm = 12 cm
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.

Answer:

i) False.
In ΔABC in which ∠B = 90º,
AB = 3, BC = 4 and AC = 5
Value of tan A = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as
it will follow the Pythagoras theorem.
AC² = AB² + BC²
5² = 3² + 4²
25 = 9 + 16
25 = 25

(ii) True.
Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
AC² = AB² + BC²
(12k)² = (5k)² + BC²
BC² + 25k² = 144k²
BC² = 119k²
Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) False.
Abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.

(iv) False.
cot A is not the product of cot and A. It is the cotangent of ∠A.

(v) False.
sin θ = Height/Hypotenuse
We know that in a right angled triangle, Hypotenuse is the longest side.
∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

Exercise 8.2

1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan²45° + cos²30° – sin²60°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos²60° + 4sec²30° – tan²45°)/(sin²30° + cos²30°)

Answer:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan²45° + cos²30° – sin²60°

(iii) cos 45°/(sec 30° + cosec 30°)

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

(v) (5cos²60° + 4sec²30° – tan²45°)/(sin²30° + cos²30°)

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan²30° =
(A) sin 60°   (B) cos 60°  (C) tan 60°  (D) sin 30°
(ii) 1-tan²45°/1+tan²45° =
(A) tan 90°   (B) 1        (C) sin 45°      (D) 0
(iii)  sin ²A = 2 sin A is true when A =
(A) 0°        (B) 30°      (C) 45°              (D) 60°
(iv) 2tan30°/1-tan²30° =
(A) cos 60°   (B) sin 60°  (C) tan 60°  (D) sin 30°

Answer:

(i) 2tan 30°/1+tan²30° =
(A) sin 60°   (B) cos 60°  (C) tan 60°  (D) sin 30°

= 

(ii) 1-tan²45°/1+tan²45° =

(A) tan 90°   (B) 1        (C) sin 45°      (D) 0

(iii) sin ²A = 2 sin A is true when A =

(A) 0°        (B) 30°      (C) 45°              (D) 60°

(iv) 2tan30°/1-tan²30° =

(A) cos 60°   (B) sin 60°  (C) tan 60°  (D) sin 30°

3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.

Answer:

……….(i)

 4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.

Answer:

(i) False.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°
= 1/2 + √3/2 = 1+√3/2

(ii) True.
sin 0° = 0
sin 30° = 1/2
sin 45° = 1/√2
sin 60° = √3/2
sin  90° = 1
Thus the value of sin θ increases as θ increases.

(iii) False.
cos 0° = 1
cos 30° = √3/2
cos 45° = 1/√2
cos 60° = 1/2
cos 90° = 0
Thus the value of cos θ decreases as θ increases.

(iv) True.
cot A = cos A/sin A
cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Exercise 8.3

1. Evaluate :
(i) sin 18°/cos 72°        (ii) tan 26°/cot 64°        (iii)  cos 48° – sin 42°       (iv)  cosec 31° – sec 59°

Answer:

(i) sin 18°/cos 72°
= sin (90° – 18°) /cos 72°
= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°
= tan (90° – 36°)/cot 64°
= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59°
= sec 59° – sec 59° = 0

2.  Show that :
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Answer:
(i) tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Answer:
tan 2A = cot (A- 18°)
⇒ cot (90° – 2A) = cot (A -18°)
Equating angles,
⇒ 90° – 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°

4.  If tan A = cot B, prove that A + B = 90°.

Answer:
tan A = cot B
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Answer:
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)
Equating angles,
90° – 4A= A- 20°
⇒ 110° = 5A
⇒ A = 22°

6. If A, B and C are interior angles of a triangle ABC, then show that
sin (B+C/2) = cos A/2

Answer:

 

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer:

sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

EXERCISE 8.4

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

 

 

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer:

 

3. Evaluate :

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Answer:

 (i) (sin²63° + sin²27°)/(cos²17° + cos²73°)

(ii)  sin 25° cos 65° + cos 25° sin 65°

Choose the correct option. Justify your choice.

4. (i) 9 sec²A – 9 tan²A =
(A) 1                 (B) 9        (C) 8                (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0                 (B) 1        (C) 2                (D) – 1
(iii) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA   (C) cosecA      (D) cosA
(iv) 1+tan²A/1+cot²A =
(A) sec²A          (B) -1      (C) cot²A                (D) tan²A

Answer:

(i)  (i) 9 sec²A – 9 tan²A =

(A) 1                 (B) 9        (C) 8                (D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0                 (B) 1        (C) 2                (D) – 1

 

(iii) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA   (C) cosecA      (D) cosA

(iv) 1+tan²A/1+cot²A =
(A) sec²A          (B) -1      (C) cot²A                (D) tan²A

 

5. Prove the following identities, where the angles involved are acute angles for which the

expressions are defined.
(i) (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin²A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec²A = 1+cot²A.

(vi) √1 + sin A/1 – sin A = sec A+ tan A

(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A

Answer:

(i) (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

[Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin²A/(1-cos A)

[Hint : Simplify LHS and RHS separately]

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec²A = 1+cot²A

(vi) √1 + sin A/1 – sin A = sec A+ tan A

LHS = 1 + sin A/(1 – sin A) …..(1)

Multiplying and dividing by (1 + sin A)

⇒ (1 + sin A)(1 + sin A/1 – sin A)(1 + sin A)

= (1 + sin A)²/(1 – sin² A) [a² – b² = (a – b)(a + b)]

= (1 + sinA)/1 – sin² A

= 1 + sin A/cos² A

= 1 + sin A/cos A

= 1/cos A + sin A/cos A

= sec A + tan A

= R.H.S

(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

[Hint : Simplify LHS and RHS separately]

(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² = tan²A