Introduction to Trigonometry

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NCERT Solutions for Chapter 8 Introduction to Trigonometry Class 10 Maths

Book Solutions

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


Exercise 8.1 Page Number 181

2

In the figure, find tan P − cot R.

Answer

In the right ∆ PQR,

Using the Pythagoras theorem, we get

Exercise 8.1 Page Number 181

3

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

Answer

Let us consider, the right ∆ ABC, we have
Perpendicular = BC

and, Hypotenuse = AC

Exercise 8.1 Page Number 181

4

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

Answer

Let in the right ∆ABC , we have
Exercise 8.1 Page Number 181

5

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

Answer


Exercise 8.1 Page Number 181

6

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

Answer

Exercise 8.1 Page Number 181

7

If cot θ =7/8, Evaluate : 

(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)

(ii) cot2θ

Answer


Exercise 8.1 Page Number 181

8

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

Answer


Exercise 8.1 Page Number 181

9

In triangle ABC, right-angled at B, if tan A = 1/√3,  find the value of :
(i) sin A cos C + cos A sinC
(ii)cos A cos C –sin A sin C

Answer

Let us consider a right ∆ABC, in which ∠B = 90°

For ∠A , we have
Base = AB
Perpendicular=BC
Hypotenuse = AC


Exercise 8.1 Page Number 181

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.

Answer

It is given that PQR is a right ∆, such that  ∠Q = 90o
PQ+QR = 25 cm
And PQ = 5cm
Let
QR = X cm
∴PR = (25−X)
∴ By Pythagoras theorem, we have
Exercise 8.1 Page Number 181

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. cosine A is abbreviated as cos A

(iv) False. cot A is a single and meaningful trigonometric ratio.

(v) False. The maximum value for sin is 1. 4/3 is greater than 1.
Exercise 8.1 Page Number 181

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

 


 

 

Exercise 8.2 Page Number 187

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

 

 

 

Exercise 8.2 Page Number 187

3

If tan(A+B) = 3and tan(A−B) = 1/√3; 0° B , find A and B.

Answer

From the table, we have tan 600 = √3  …(1)
Also tan (A+B) = √3  (Given) ….(2)
From (1) and (2) , we get
A+B = 60° …. (3)
Similarly,
A−B = 30° …. (4)
Adding (3) and (4) ,
2A =90°
⇒A = 45°
Subtracting (4) from (3) , we get

2B = 30°

⇒ B = 15°

Exercise 8.2 Page Number 187

4

State whether the foolowing are true or false. Justify your answer.
(i) sin (A+B)= sin A +sin B .
(ii)The value of sin

Answer

(i) Let us take A = 30° and B= 60°
then LHS = sin(30°+ 60°)
= sin90° = 1
RHS = sin30°+sin60°

Since, LHS ≠ RHS
∴The statement sin(A+B) = sin A+ sin B is false .

 

 

 

 

(v) From the table, we have,
cot0° = not defined.
∴ The given statement is true.

Exercise 8.2 Page Number 187

1

Evaluate :

(i) sin 18°/cos 72°

(ii) tan 26°/cot 64°

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

Answer

 


(iii) cos 48°− sin 42° 
∵ cos 48° = cos(90°− 42°) = sin 42°  [∵ cos (90°−A) = sin A ] 
∴ cos 48°− sin 42°  =  sin 42° − sin 452°  = 0

 

(iv) cosec 31° – sec 59° 
∵ cosec 31°  = cosec (90° − 59° ) = sec 59°  [∵ cosec (90°−A) = sec A ]
∴ cosec 31° – sec 59° = sec 59°− sec 59° = 0.

Exercise 8.3 Page Number 189

2

Show that :
(i) tan48°  tan23° tan42° tan67° = 1
(ii) cos 38° cos52° – sin38° sin52° = 0

Answer

(i) tan 48° tan23° tan42° tan67° = 1
L.H.S. =  tan 48° tan 23° tan42° tan67° 
= tan(90° − 42°) tan 23° tan42° tan(90° − 23°)
= cot 42°  tan 23°  tan42°  cot 23° 

= R.H.S.
Thus ,  tan 48° tan23° tan42° tan67° = 1 

 

(ii) cos 38° cos 52°– sin 38° sin 52° = 0
L.H.S. =mcos 38° cos 52°– sin 38° sin 52°  
= cos 38°  cos(90° − 38°)  − sin38°  sin(90° −38°)
= cos 38°   sin 38°  − sin 38° cos 38°   
[∵ sin(90° – A) = cos A and  cos(90° – A ) = sin A )
= 0 = R.H.S.
Thus ,  cos 38° cos 52° – sin 38° sin 52° = 0

Exercise 8.3 Page Number 189

3

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

Answer

tan 2A = cot( A− 18° )
 Also tan (2A)° = cot (90° – 2A)  [∵  tan θ = cot(90° – θ)]
⇒ A − 18° = 90° – 2A
⇒ A +2A =  90°  + 180
⇒ 3A = 108°
Exercise 8.3 Page Number 189

4

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

Answer

tan A = cot B  (Given)
and cot B = tan(90° –B)    [∵ tan(90°−θ) = cot θ]
∴ A = 90° –B
⇒ A+B = 90°.
Exercise 8.3 Page Number 189

5

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

Answer

∵ sec4A = cosec (A− 20°)
And sec 4A = cosec (90°− 4A) [∵ cosec (90°− θ) = sec θ]
∴ A− 20°= 90°− 4A 
⇒ A +4A = 90°+ 20° 
⇒ 5A = 110°
Exercise 8.3 Page Number 189

6

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

Answer

Since , sum of the angles of ∆ ABC is A°+B°+C° = 180°
∴B+C = 180° –A
Dividing both sides by 2 ,
Exercise 8.3 Page Number 190

7

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

Answer

Since sin 67° =  sin (90°− 23°)
= cos 23°  [∵ sin (90°− θ)= cos θ ]
Also,  cos 75° = cos (90° − 15°)
= Sin 15° [∵cos (90°−θ)= sin θ]
∴ We have :
sin 67°+ cos75°= cos23°  + sin15°. 
Exercise 8.3 Page Number 190

1

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

Answer


Exercise 8.4 Page Number 193

2

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

Answer


Exercise 8.4 Page Number 193

3

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

Answer


 

(ii) sin25° cos 65° + cos25° sin65°
∵  sin 25°= sin (90°−65°) = cos65°  [∵ sin (90°−A) = cos A]
And cos 25° = cos (90°− 65°) = sin 65°  [∵ cos (90°− A) = sin A]
∴ sin 25°cos 65° + cos 25° sin 65°
= cos 65°  cos 65°  + sin 65°  sin 65°
= cos265° + sin2 65°  [∵ cos2A+ sin2A= 1]
= 1

Exercise 8.4 Page Number 193

4

Choose the correct option. Justify your choice.
(i)  9 sec2A− 9 tan2A = ….
(A) 1
(B) 0
(C)8
(D) 0


(ii) (1+ tan

Answer

(i) Since , 9 sec2A− 9 tan2A
= 9(sec2A− tan2A )
= 9(1)  [∵  tan2 A +1 =  sec2A ⇒  sec2A−  tan2 A  = 1 ]
= 9




Exercise 8.4 Page Number 193

5(i)

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)

Answer

Exercise 8.4 Page Number 193

5(ii)

Prove: cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

Answer

Exercise 8.4 Page Number 194

5(iii)

Prove: 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]

Answer

Exercise 8.4 Page Number 194

5(iv)

Prove: (1 + sec A)/sec A = sin2A/(1-cos A)

[Hint : Simplify LHS and RHS separately]

Answer

Exercise 8.4 Page Number 194

5(v)

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

Answer

Exercise 8.4 Page Number 194

5(vi)

Prove: 

Answer

Exercise 8.4 Page Number 194

5(vii)

Prove: (sin θ - 2sin3θ)/(2cos3θ-cos θ) = tan θ

Answer

Exercise 8.4 Page Number 194

5(viii)

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

Answer

Exercise 8.4 Page Number 194

5(ix)

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

[Hint : Simplify LHS and RHS separately]

Answer

Exercise 8.4 Page Number 194

5(x)

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

Answer

Exercise 8.4 Page Number 194

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