Bihar Board Class 10th Maths Chapter 8 Introduction to Trigonometry Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ∆ ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A (ii) sin C, cos C
Solution:
Let us draw a right ∆ ABC.

By using the Pythagoras theorem, we have :
AC² = AB² + BC² = (24)² + (7)²
= 576 + 49 = 625
So, AC = 625−−−√ cm = 25 cm
(i) sin A = BCAC = 725, cos A = ABAC = 2425
(ii) sin C = ABAC = 2425, cos C = BCAC = 725

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

Solution:
By using the Pythagoras theorem, we have :
PR² = PQ² + QR² or 13² = 12² + QR²
or QR² = 13² – 12² = 169 – 144 = 25
So, QR = 25−−√ cm = 5 cm
∴ tan P = QPPQ = 512 and cot R = QRPQ = 512
Hence, tan P – cot R = 512 – 512 = 0.

Question 3.
If sin A = 34, calculate cos A and tan A.
Solution:
Consider a ∆ ABC in which ∠B = 90°.

For ∠A, we have :
Base (adjacent side) = AB,
Perp. (opposite side) = BC
and Hyp. = AC.

Question 4.
Given 15 cot A =8, find sin A and sec A.
Solution:
Consider a ∆ ABC in which ∠B = 90°.
For ∠A, we have:
Base AB, Perp. = BC and Hyp. = AC.

Question 5.
Given sec θ = 1312, calculate all other trigonometric ratios.
Solution:
Consider a ∆ ABC in which ∠A = 0° and ∠B = 90° Then, Base = AB, Perp. = BC and Hyp. = AC.

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let us consider two right triangles PQA and RSB in which cos A = cos B (see the figure).

We have:
cos A = QAPA
and cos B = SBRB
Thus, it is given that
QAPA = SBRB
so, QASB = PARB = K(say) … (1)
Now, by Pythagoras Theorem,

Therefore, from (1) and (2), we have
QASB = PARB = PQRS
Hence, ∆ PQA ~ ∆ RSB [SSS similarity]
Therefore, ∠A = ∠B [Corresponding angles]

Question 7.
If cot θ = 78, evaluate :
(i) (1+sinθ)(1−sinθ)
(ii) cot² θ
Solution:
Consider a ∆ ABC in which ∠A = 0 and ∠B = 90°.

Then, Base = AB, Perp = BC and Hyp = AC
∴ cot θ = BasePerp = ABBC = 78
Let AB = 7k and BC = 8k

Question 8.
If 3 cot A = 4, check whether 1−tan2A = cos² A – sin² A or not.
Solution:
Consider a ∆ ABC in which ∠B = 90°.
Foe ∠A, we have:

Base = AB, Perp = BC and Hyp = AC
∴ cot A = BasePerp = ABBC = 43
Let AB = 4k and BC = 3k [3 cot A = 4 ⇒ cot A = 43 ]

Question 9.
In ∆ ABC right angled at B, if tan A = 13√, find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
Consider a ∆ ABC, in which ∠B = 90°

For ∠A, we have :
Base = AB,
Perp. = BC .
and Hyp. = AC
∴ tan A = PerpBase = BCAB = 13√
Let BC = k and AB = 3–√k.

(i) sin A cos C + cos A sin C = 12 x 12 + 3√2 x 3√2
= 14 x 34 = 44 = 1
(ii) cos A cos C – sin A sin C = 3√2 x 12 – 12 3√2 = 0

Question 10
In ∆ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
In ∆ PQR, right ∠d at Q,

PR + QR = 25 cm and PQ = 5 cm.
Let QR = x cm
∴ PR = (25 – x) cm
By Pythagoras theorem, we have :
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RP² = RQ² + QP²
So, (25 – x)² = x² + 5²
or 625 – 50x + x² = x² + 25
or – 50x = – 600 or x = −600−50 = 12
∴ RQ = 12cm
So, RP = (25 – 12)cm = 13 cm
Now, sin P = RQRP = 1213
cos P = PQRP = 513
and, tan P = RQPQ = 125

Question 11.
State whether the following are true or false. Justify your answer.

1. The value of tan A is always less than 1.
2. sec A = 125 for some value of angle A.
3. cos A is the abbreviation used for the cosecant of angle A.
4. cot A is the product of cot and A.
5. sin θ = 43 for some angle θ.

Solution:

1. False because sides of a right triangle may have any length, so tan A may have any value.
2. True as sec A is always greater than or equal to 1.
3. False as cos A is the abbreviation used for cosine A.
4. False as cot A is not the product of ‘cot’ and A. ‘cot separated from A has no meaning.
5. False as sin 0 cannot be > 1.

BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°

Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
= 3√2 x 3√2 + 12 x 12
= 34 + 14 = 3+14 = 44 = 1.

(ii) 2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + (3√2)2 – (3√2)2
= 2 + 34 – 34 = 2

Question 2.
Choose the correct option and justify:
(i) 2tan30∘
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) 1−tan245∘
(A) tan 90°
(B) 1
(C) sin 45°
(D) O

(iii) sin 2A = 2 sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) 2tan30∘
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) none of these
Solution:
(i) (A)

(ii) (D)
Because 1−tan245∘ = 1−11+1 = 02 = 0

(iii) (A)
Because when A – 0°, sin 2A = sin 0° = 0
and, 2 sin A = 2 sin 0°= 2 x 0 = 0
or sin 2 A = 2 sin A, when A = 0°

(iv) (C)

Question 3.
If tan (A + B) = 3–√ and tan (A – B) = 13√; 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
tan (A + B) = 3–√ means tan (A + B) = tan 60°
So, A + B = 60° … (1)
tan (A – B) = 13√ means tan (A – B) = tan 30°
So, A – B = 30° … (2)
Solving (1) and (2), we get
A = 45° and B = 15°
Hence, A = 45° and B = 15°.

Question 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 0 = cos 0 for all values of 0.
(v) cot A is not defined for A = 0°.
Solution:
(i) False. Because
When A = 60° and B = 30°. Then,
sin (A + B) = sin (60° + 30°) = sin 90° = 1
and, sin A + sin B = sin 60° + sin 30°
= 3√2 + 12 = 3√+12
So, sin (A + B) ≠ sin A + sin B

(ii) True. Because, it is clear from the table below :

that the value of sin θ increases as θ increases.

(iii) False. Because it is clear from the table below :

that the value of cos θ decreases as θ increases.

(iv) False. Because it is only true for θ = 45°.
(sin 45° = 12√ = cos 45°)

(v) True. Because tan 0° = 0 and
cot 0° = tan0∘ = 10, i.e., not defined.

BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
Evaluate :
(i) sin18∘
(ii) tan26∘
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Solution:
(i) sin18∘
= sin(90∘−72∘)
= cos72∘
= 1

(ii) tan26∘
= tan(90∘−64∘)
= cot64∘
= 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

Question 2.
Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(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°
= tan42∘.tan67∘.tan 42° tan 67° = 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

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
We are given that
tan 2A = cot (A – 18°) … (1)
Since tan 2A = cot (90° – 2A), so we can write (1) as
cot (90° – 2A) = cot (A – 18°)
Since (90° – 2A) and (A – 18°) are both acute angles, therefore
90° – 2A = A – 18°
or – 2A – A = – 18° – 90° or – 3A = – 108°
or A = 36°

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
We are given that
tan A = cot B … (1)
Since tan A = cot (90° – A), so we can write (1)
as cot (90° – A) = cot B
Since (90° – A) and B are both acute angles, therefore
90° – A = B or A + B = 90°

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
We are given that
sec 4A = cosec (A – 20°). …(1)
Since sec 4A = cosec (90° – 4A), so we can write (1) as
cosec (90° – 4A) = cosec (A – 20°)
Since (90° – 4A) and (A – 20°) are both acute angles, therefore
90° – 4A = A – 20° or – 4A – A = – 20° – 90°
or – 5A = – 110° or A = 22°

Question 6.
If A, B and C are, interior angles of a ∆ ABC, then show, that
sin(B+C2) = cosA2
Solution:
Since A, B and C are the interior angles of a ∆ ABC, therefore

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
Consider a ∆ ABC, in which ∠B = 90°.

Base = AB
Perp. = BC
and Hyp. = AC
∴ cot A = \(\frac{\text { Base }}{\text { Perp }}\) = \(\frac { AB }{ BC }\)
or \(\frac { AB }{ BC }\) = cot A = \(\frac { cot A }{ 1 }\)
Let AB = k cot A and BC = k.

Question 2.
Write the other trigonometric ratios of A in terms of sec A.
Solution:
Consider a ∆ ABC, in which ∠B = 90°

For ∠A, we have :
Base = AB,
Perp = BC
Hyp = AC.
∴ sec A = \(\frac{\text { Hyp }}{\text { Base }}\) = \(\frac { AC }{ AB }\)
or \(\frac { AC }{ AB }\) = sec A = \(\frac { sec A }{ 1 }\)
Let AB = k and AC = k sec A.

Question 3.
Evaluate :
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
(ii) sin 25° cos 65° + cos 25° sin 65
Solution:
(i) Here, sin 63° = sin (90° – 27°) = cos 27°
and cos 17° = cos (90° – 73°) = sin 73°

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin (90° – 65°) cos 65° + cos (90° – 65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos² 65° + sin² 65° = 1

Question 4.
Choose the correct option. Justify your choice :
(i) 9 sec² A – 9 tan² A =
(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) none of these
(iii) (sec A + tan A)(1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)
(A) sec² A
(B) – 1
(C) cot² A
(D) none of these
Solution:
(i) (B), because
9 sec² A – 9 tan² A = 9 (sec² A – tan² A) = 9 x 1 = 9

(ii) (C), because
(1 + tan θ + sec θ )(1 + cot θ – cosec θ)

(iii) (D), because
(sec A + tan A)(1 – sin A) =

(iv) (D), because

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined :

Solution:
(i) We have :
L.H.S. = (cosec θ – cot θ)²

(ii) We have :

(ii) We have :

(iii) We have :

(iv) We have :

(v) We have :

(vi) We have :

(vii) We have :

(viii) We have :
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A cosec A) + (cos² A + sec2 A + 2 cos A sec A)
= (sin² A + cosec² A + 2 sin A.\(\frac{1}{\sin A}\)) + (cos² A + sec2 A + 2 cos A. \(\frac{1}{\cos A}\))
= (sin²A + cosec² A + 2) + (cos² A + sec² A + 2)
= sin²A + cos² A + cosec² A + sec² A + 4
= 1 + (1 + cot² θ) + (1 + tan² A) + 7 + tan² A + cot² A [ ∵ cosec² A = 1 + cot² A and sec² A = 1 + tan² A]
= R.H.S.

(ix) We have :

(x) We have :