Class 10: Maths Chapter 5 solutions. Complete Class 10 Maths Chapter 5 Notes.
Contents
RD Sharma Solutions for Class 10 Maths Chapter 5–Trigonometric Ratios
RD Sharma 10th Maths Chapter 5, Class 10 Maths Chapter 5 solutions
Exercise 5.1 Page No: 5.23
1. In each of the following, one of the six trigonometric ratios s given. Find the values of the other trigonometric ratios.
(i) sin A = 2/3
Solution:
We have,
sin A = 2/3 ……..….. (1)
As we know, by sin definition;
sin A = Perpendicular/ Hypotenuse = 2/3 ….(2)
By comparing eq. (1) and (2), we have
Opposite side = 2 and Hypotenuse = 3
Now, on using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get
⇒ 32 = AB2 + 22
AB2 = 32 – 22
AB2 = 9 – 4
AB2 = 5
AB = √5
Hence, Base = √5
By definition,
cos A = Base/Hypotenuse
⇒ cos A = √5/3
Since, cosec A = 1/sin A = Hypotenuse/Perpendicular
⇒ cosec A = 3/2
And, sec A = Hypotenuse/Base
⇒ sec A = 3/√5
And, tan A = Perpendicular/Base
⇒ tan A = 2/√5
And, cot A = 1/ tan A = Base/Perpendicular
⇒ cot A = √5/2
(ii) cos A = 4/5
Solution:
We have,
cos A = 4/5 …….…. (1)
As we know, by cos defination
cos A = Base/Hypotenuse …. (2)
By comparing eq. (1) and (2), we get
Base = 4 and Hypotenuse = 5
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of base (AB) and hypotenuse (AC) and for the perpendicular (BC), we get
52 = 42 + BC2
BC2 = 52 – 42
BC2 = 25 – 16
BC2 = 9
BC= 3
Hence, Perpendicular = 3
By definition,
sin A = Perpendicular/Hypotenuse
⇒ sin A = 3/5
Then, cosec A = 1/sin A
⇒ cosec A= 1/ (3/5) = 5/3 = Hypotenuse/Perependicular
And, sec A = 1/cos A
⇒ sec A =Hypotenuse/Base
sec A = 5/4
And, tan A = Perpendicular/Base
⇒ tan A = 3/4
Next, cot A = 1/tan A = Base/Perpendicular
∴ cot A = 4/3
(iii) tan θ = 11/1
Solution:
We have, tan θ = 11…..…. (1)
By definition,
tan θ = Perpendicular/ Base…. (2)
On Comparing eq. (1) and (2), we get;
Base = 1 and Perpendicular = 5
Now, using Pythagoras theorem in Δ ABC.
AC2 = AB2 + BC2
Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get;
AC2 = 12 + 112
AC2 = 1 + 121
AC2= 122
AC= √122
Hence, hypotenuse = √122
By definition,
sin = Perpendicular/Hypotenuse
⇒ sin θ = 11/√122
And, cosec θ = 1/sin θ
⇒ cosec θ = √122/11
Next, cos θ = Base/ Hypotenuse
⇒ cos θ = 1/√122
And, sec θ = 1/cos θ
⇒ sec θ = √122/1 = √122
And, cot θ = 1/tan θ
∴ cot θ = 1/11
(iv) sin θ = 11/15
Solution:
We have, sin θ = 11/15 ………. (1)
By definition,
sin θ = Perpendicular/ Hypotenuse …. (2)
On Comparing eq. (1) and (2), we get;
Perpendicular = 11 and Hypotenuse= 15
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have
152 = AB2 +112
AB2 = 152 – 112
AB2 = 225 – 121
AB2 = 104
AB = √104
AB= √ (2×2×2×13)
AB= 2√(2×13)
AB= 2√26
Hence, Base = 2√26
By definition,
cos θ = Base/Hypotenuse
∴ cosθ = 2√26/ 15
And, cosec θ = 1/sin θ
∴ cosec θ = 15/11
And, secθ = Hypotenuse/Base
∴ secθ =15/ 2√26
And, tan θ = Perpendicular/Base
∴ tanθ =11/ 2√26
And, cot θ = Base/Perpendicular
∴ cotθ =2√26/ 11
(v) tan α = 5/12
Solution:
We have, tan α = 5/12 …. (1)
By definition,
tan α = Perpendicular/Base…. (2)
On Comparing eq. (1) and (2), we get
Base = 12 and Perpendicular side = 5
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of base (AB) and the perpendicular (BC) to get hypotenuse (AC), we have
AC2 = 122 + 52
AC2 = 144 + 25
AC2= 169
AC = 13 [After taking sq root on both sides]
Hence, Hypotenuse = 13
By definition,
sin α = Perpendicular/Hypotenuse
∴ sin α = 5/13
And, cosec α = Hypotenuse/Perpendicular
∴ cosec α = 13/5
And, cos α = Base/Hypotenuse
∴ cos α = 12/13
And, sec α =1/cos α
∴ sec α = 13/12
And, tan α = sin α/cos α
∴ tan α=5/12
Since, cot α = 1/tan α
∴ cot α =12/5
(vi) sin θ = √3/2
Solution:
We have, sin θ = √3/2 …………. (1)
By definition,
sin θ = Perpendicular/ Hypotenuse….(2)
On Comparing eq. (1) and (2), we get;
Perpendicular = √3 and Hypotenuse = 2
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) and get the base (AB), we get;
22 = AB2 + (√3)2
AB2 = 22 – (√3)2
AB2 = 4 – 3
AB2 = 1
AB = 1
Thus, Base = 1
By definition,
cos θ = Base/Hypotenuse
∴ cos θ = 1/2
And, cosec θ = 1/sin θ
Or cosec θ= Hypotenuse/Perpendicualar
∴ cosec θ =2/√3
And, sec θ = Hypotenuse/Base
∴ sec θ = 2/1
And, tan θ = Perpendicula/Base
∴ tan θ = √3/1
And, cot θ = Base/Perpendicular
∴ cot θ = 1/√3
(vii) cos θ = 7/25
Solution:
We have, cos θ = 7/25 ……….. (1)
By definition,
cos θ = Base/Hypotenuse
On Comparing eq. (1) and (2), we get;
Base = 7 and Hypotenuse = 25
Now, using Pythagoras theorem in Δ ABC
AC2= AB2 + BC2
Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC)
252 = 72 +BC2
BC2 = 252 – 72
BC2 = 625 – 49
BC2 = 576
BC= √576
BC= 24
Hence, Perpendicular side = 24
By definition,
sin θ = perpendicular/Hypotenuse
∴ sin θ = 24/25
Since, cosec θ = 1/sin θ
Also, cosec θ= Hypotenuse/Perpendicualar
∴ cosec θ = 25/24
Since, sec θ = 1/cosec θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 25/7
Since, tan θ = Perpendicular/Base
∴ tan θ = 24/7
Now, cot = 1/tan θ
So, cot θ = Base/Perpendicular
∴ cot θ = 7/24
(viii) tan θ = 8/15
Solution:
We have, tan θ = 8/15 …………. (1)
By definition,
tan θ = Perpendicular/Base …. (2)
On Comparing eq. (1) and (2), we get;
Base = 15 and Perpendicular = 8
Now, using Pythagoras theorem in Δ ABC
AC2 = 152 + 82
AC2 = 225 + 64
AC2 = 289
AC = √289
AC = 17
Hence, Hypotenuse = 17
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ = 8/17
Since, cosec θ = 1/sin θ
Also, cosec θ = Hypotenuse/Perpendicualar
∴ cosec θ = 17/8
Since, cos θ = Base/Hypotenuse
∴ cos θ = 15/17
Since, sec θ = 1/cos θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 17/15
Since, cot θ = 1/tan θ
Also, cot θ = Base/Perpendicular
∴ cot θ = 15/8
(ix) cot θ = 12/5
Solution:
We have, cot θ = 12/5 …………. (1)
By definition,
cot θ = 1/tan θ
cot θ = Base/Perpendicular ……. (2)
On Comparing eq. (1) and (2), we have
Base = 12 and Perpendicular side = 5
Now, using Pythagoras theorem in Δ ABC
AC2= AB2 + BC2
Putting the value of base (AB) and perpendicular (BC) to get the hypotenuse (AC);
AC2 = 122 + 52
AC2= 144 + 25
AC2 = 169
AC = √169
AC = 13
Hence, Hypotenuse = 13
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ= 5/13
Since, cosec θ = 1/sin θ
Also, cosec θ= Hypotenuse/Perpendicualar
∴ cosec θ = 13/5
Since, cos θ = Base/Hypotenuse
∴ cos θ = 12/13
Since, sec θ = 1/cosθ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 13/12
Since, tanθ = 1/cot θ
Also, tan θ = Perpendicular/Base
∴ tan θ = 5/12
(x) sec θ = 13/5
Solution:
We have, sec θ = 13/5…….… (1)
By definition,
sec θ = Hypotenuse/Base…………. (2)
On Comparing eq. (1) and (2), we get
Base = 5 and Hypotenuse = 13
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
And. putting the value of base side (AB) and hypotenuse (AC) to get the perpendicular side (BC)
132 = 52 + BC2
BC2 = 132 – 52
BC2=169 – 25
BC2= 144
BC= √144
BC = 12
Hence, Perpendicular = 12
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ= 12/13
Since, cosec θ= 1/ sin θ
Also, cosec θ= Hypotenuse/Perpendicualar
∴ cosec θ = 13/12
Since, cos θ= 1/sec θ
Also, cos θ = Base/Hypotenuse
∴ cos θ = 5/13
Since, tan θ = Perpendicular/Base
∴ tan θ = 12/5
Since, cot θ = 1/tan θ
Also, cot θ = Base/Perpendicular
∴ cot θ = 5/12
(xi) cosec θ = √10
Solution:
We have, cosec θ = √10/1 ……..… (1)
By definition,
cosec θ = Hypotenuse/ Perpendicualar …….….(2)
And, cosecθ = 1/sin θ
On comparing eq.(1) and(2), we get
Perpendicular side = 1 and Hypotenuse = √10
Now, using Pythagoras theorem in Δ ABC
AC2 = AB2 + BC2
Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base side (AB)
(√10)2 = AB2 + 12
AB2= (√10)2 – 12
AB2= 10 – 1
AB = √9
AB = 3
So, Base side = 3
By definition,
Since, sin θ = Perpendicular/Hypotenuse
∴ sin θ = 1/√10
Since, cos θ = Base/Hypotenuse
∴ cos θ = 3/√10
Since, sec θ = 1/cos θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = √10/3
Since, tan θ = Perpendicular/Base
∴ tan θ = 1/3
Since, cot θ = 1/tan θ
∴ cot θ = 3/1
(xii) cos θ =12/15
Solution:
We have; cos θ = 12/15 ………. (1)
By definition,
cos θ = Base/Hypotenuse……… (2)
By comparing eq. (1) and (2), we get;
Base =12 and Hypotenuse = 15
Now, using Pythagoras theorem in Δ ABC, we get
AC2 = AB2+ BC2
Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC);
152 = 122 + BC2
BC2 = 152 – 122
BC2 = 225 – 144
BC 2= 81
BC = √81
BC = 9
So, Perpendicular = 9
By definition,
Since, sin θ = perpendicular/Hypotenuse
∴ sin θ = 9/15 = 3/5
Since, cosec θ = 1/sin θ
Also, cosec θ = Hypotenuse/Perpendicualar
∴ cosec θ= 15/9 = 5/3
Since, sec θ = 1/cos θ
Also, sec θ = Hypotenuse/Base
∴ sec θ = 15/12 = 5/4
Since, tan θ = Perpendicular/Base
∴ tan θ = 9/12 = 3/4
Since, cot θ = 1/tan θ
Also, cot θ = Base/Perpendicular
∴ cot θ = 12/9 = 4/3
2. In a △ ABC, right angled at B, AB = 24 cm , BC = 7 cm. Determine
(i) sin A , cos A (ii) sin C, cos C
Solution:
(i) Given: In △ABC, AB = 24 cm, BC = 7cm and ∠ABC = 90o
To find: sin A, cos A
By using Pythagoras theorem in △ABC we have
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC2= 625
AC = √625
AC= 25
Hence, Hypotenuse = 25
By definition,
sin A = Perpendicular side opposite to angle A/ Hypotenuse
sin A = BC/ AC
sin A = 7/ 25
And,
cos A = Base side adjacent to angle A/Hypotenuse
cos A = AB/ AC
cos A = 24/ 25
(ii) Given: In △ABC , AB = 24 cm and BC = 7cm and ∠ABC = 90o
To find: sin C, cos C
By using Pythagoras theorem in △ABC we have
AC2 = AB2 + BC2
AC2 = 242 + 72
AC2 = 576 + 49
AC2= 625
AC = √625
AC= 25
Hence, Hypotenuse = 25
By definition,
sin C = Perpendicular side opposite to angle C/Hypotenuse
sin C = AB/ AC
sin C = 24/ 25
And,
cos C = Base side adjacent to angle C/Hypotenuse
cos A = BC/AC
cos A = 7/25
3. In fig. 5.37, find tan P and cot R. Is tan P = cot R?
Solution:
By using Pythagoras theorem in △PQR, we have
PR2 = PQ2 + QR2
Putting the length of given side PR and PQ in the above equation
132 = 122 + QR2
QR2 = 132 – 122
QR2 = 169 – 144
QR2 = 25
QR = √25 = 5
By definition,
tan P = Perpendicular side opposite to P/ Base side adjacent to angle P
tan P = QR/PQ
tan P = 5/12 ………. (1)
And,
cot R= Base/Perpendicular
cot R= QR/PQ
cot R= 5/12 …. (2)
When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.
Therefore, L.H.S of both equations should also be equal.
∴ tan P = cot R
Yes, tan P = cot R = 5/12
4. If sin A = 9/41, compute cos A and tan A.
Solution:
Given that, sin A = 9/41 …………. (1)
Required to find: cos A, tan A
By definition, we know that
sin A = Perpendicular/ Hypotenuse……………(2)
On Comparing eq. (1) and (2), we get;
Perpendicular side = 9 and Hypotenuse = 41
Let’s construct △ABC as shown below,
And, here the length of base AB is unknown.
Thus, by using Pythagoras theorem in △ABC, we get;
AC2 = AB2 + BC2
412 = AB2 + 92
AB2 = 412 – 92
AB2 = 168 – 81
AB= 1600
AB = √1600
AB = 40
⇒ Base of triangle ABC, AB = 40
We know that,
cos A = Base/ Hypotenuse
cos A =AB/AC
cos A =40/41
And,
tan A = Perpendicular/ Base
tan A = BC/AB
tan A = 9/40
5. Given 15cot A= 8, find sin A and sec A.
Solution
We have, 15cot A = 8
Required to find: sin A and sec A
As, 15 cot A = 8
⇒ cot A = 8/15 …….(1)
And we know,
cot A = 1/tan A
Also by definition,
Cot A = Base side adjacent to ∠A/ Perpendicular side opposite to ∠A …. (2)
On comparing equation (1) and (2), we get;
Base side adjacent to ∠A = 8
Perpendicular side opposite to ∠A = 15
So, by using Pythagoras theorem to △ABC, we have
AC2 = AB2 +BC2
Substituting values for sides from the figure
AC2 = 82 + 152
AC2 = 64 + 225
AC2 = 289
AC = √289
AC = 17
Therefore, hypotenuse =17
By definition,
sin A = Perpendicular/Hypotenuse
⇒ sin A= BC/AC
sin A= 15/17 (using values from the above)
Also,
sec A= 1/ cos A
⇒ secA = Hypotenuse/ Base side adjacent to ∠A
∴ sec A= 17/8
6. In △PQR, right-angled at Q, PQ = 4cm and RQ = 3 cm. Find the value of sin P, sin R, sec P and sec R.
Solution:
Given:
△PQR is right-angled at Q.
PQ = 4cm
RQ = 3cm
Required to find: sin P, sin R, sec P, sec R
Given △PQR,
By using Pythagoras theorem to △PQR, we get
PR2 = PQ2 +RQ2
Substituting the respective values,
PR2 = 42 +32
PR2 = 16 + 9
PR2 = 25
PR = √25
PR = 5
⇒ Hypotenuse =5
By definition,
sin P = Perpendicular side opposite to angle P/ Hypotenuse
sin P = RQ/ PR
⇒ sin P = 3/5
And,
sin R = Perpendicular side opposite to angle R/ Hypotenuse
sin R = PQ/ PR
⇒ sin R = 4/5
And,
sec P=1/cos P
secP = Hypotenuse/ Base side adjacent to ∠P
sec P = PR/ PQ
⇒ sec P = 5/4
Now,
sec R = 1/cos R
secR = Hypotenuse/ Base side adjacent to ∠R
sec R = PR/ RQ
⇒ sec R = 5/3
7. If cot θ = 7/8, evaluate
(i) (1+sin θ)(1–sin θ)/ (1+cos θ)(1–cos θ)
(ii) cot2 θ
Solution:
(i) Required to evaluate:
, given = cot θ = 7/8
Taking the numerator, we have
(1+sin θ)(1–sin θ) = 1 – sin2 θ [Since, (a+b)(a-b) = a2 – b2]
Similarly,
(1+cos θ)(1–cos θ) = 1 – cos2 θ
We know that,
sin2 θ + cos2 θ = 1
⇒ 1 – cos2 θ = sin2 θ
And,
1 – sin2 θ = cos2 θ
Thus,
(1+sin θ)(1 –sin θ) = 1 – sin2 θ = cos2 θ
(1+cos θ)(1–cos θ) = 1 – cos2 θ = sin2 θ
⇒
= cos2 θ/ sin2 θ
= (cos θ/sin θ)2
And, we know that (cos θ/sin θ) = cot θ
⇒
= (cot θ)2
= (7/8)2
= 49/ 64
(ii) Given,
cot θ = 7/8
So, by squaring on both sides we get
(cot θ)2 = (7/8)2
∴ cot θ2 = 49/64
8. If 3cot A = 4, check whether (1–tan2A)/(1+tan2A) = (cos2A – sin2A) or not.
Solution:
Given,
3cot A = 4
⇒ cot A = 4/3
By definition,
tan A = 1/ Cot A = 1/ (4/3)
⇒ tan A = 3/4
Thus,
Base side adjacent to ∠A = 4
Perpendicular side opposite to ∠A = 3
In ΔABC, Hypotenuse is unknown
Thus, by applying Pythagoras theorem in ΔABC
We get
AC2 = AB2 + BC2
AC2 = 42 + 32
AC2 = 16 + 9
AC2 = 25
AC = √25
AC = 5
Hence, hypotenuse = 5
Now, we can find that
sin A = opposite side to ∠A/ Hypotenuse = 3/5
And,
cos A = adjacent side to ∠A/ Hypotenuse = 4/5
Taking the LHS,
Thus, LHS = 7/25
Now, taking RHS
9. If tan θ = a/b, find the value of (cos θ + sin θ)/ (cos θ – sin θ)
Solution:
Given,
tan θ = a/b
And, we know by definition that
tan θ = opposite side/ adjacent side
Thus, by comparison
Opposite side = a and adjacent side = b
To find the hypotenuse, we know that by Pythagoras theorem that
Hypotenuse2 = opposite side2 + adjacent side2
⇒ Hypotenuse = √(a2 + b2)
So, by definition
sin θ = opposite side/ Hypotenuse
sin θ = a/ √(a2 + b2)
And,
cos θ = adjacent side/ Hypotenuse
cos θ = b/ √(a2 + b2)
Now,
After substituting for cos θ and sin θ, we have
∴
Hence Proved.
10. If 3 tan θ = 4, find the value of
Solution:
Given, 3 tan θ = 4
⇒ tan θ = 4/3
From, let’s divide the numerator and denominator by cos θ.
We get,
(4 – tan θ) / (2 + tan θ)
⇒ (4 – (4/3)) / (2 + (4/3)) [using the value of tan θ]
⇒ (12 – 4) / (6 + 4) [After taking LCM and cancelling it]
⇒ 8/10 = 4/5
∴ = 4/5
11. If 3 cot θ = 2, find the value of
Solution:
Given, 3 cot θ = 2
⇒ cot θ = 2/3
From, let’s divide the numerator and denominator by sin θ.
We get,
(4 –3 cot θ) / (2 + 6 cot θ)
⇒ (4 – 3(2/3)) / (2 + 6(2/3)) [using the value of tan θ]
⇒ (4 – 2) / (2 + 4) [After taking LCM and simplifying it]
⇒ 2/6 = 1/3
∴ = 1/3
12. If tan θ = a/b, prove that
Solution:
Given, tan θ = a/b
From LHS, let’s divide the numerator and denominator by cos θ.
And we get,
(a tan θ – b) / (a tan θ + b)
⇒ (a(a/b) – b) / (a(a/b) + b) [using the value of tan θ]
⇒ (a2 – b2)/b2 / (a2 + b2)/b2 [After taking LCM and simplifying it]
⇒ (a2 – b2)/ (a2 + b2)
= RHS
– Hence Proved
13. If sec θ = 13/5, show that
Solution:
Given,
sec θ = 13/5
We know that,
sec θ = 1/ cos θ
⇒ cos θ = 1/ sec θ = 1/ (13/5)
∴ cos θ = 5/13 ……. (1)
By definition,
cos θ = adjacent side/ hypotenuse ….. (2)
Comparing (1) and (2), we have
Adjacent side = 5 and hypotenuse = 13
By Pythagoras theorem,
Opposite side = √((hypotenuse) 2 – (adjacent side)2)
= √(132 – 52)
= √(169 – 25)
= √(144)
= 12
Thus, opposite side = 12
By definition,
tan θ = opposite side/ adjacent side
∴ tan θ = 12/ 5
From, let’s divide the numerator and denominator by cos θ.
We get,
(2 tan θ – 3) / (4 tan θ – 9)
⇒ (2(12/5) – 3) / (4(12/5) – 9) [using the value of tan θ]
⇒ (24 – 15) / (48 – 45) [After taking LCM and cancelling it]
⇒ 9/3 = 3
∴ = 3
14. If cos θ = 12/13, show that sin θ(1 – tan θ) = 35/156
Solution:
Given, cos θ = 12/13…… (1)
By definition we know that,
cos θ = Base side adjacent to ∠θ / Hypotenuse……. (2)
When comparing equation (1) and (2), we get
Base side adjacent to ∠θ = 12 and Hypotenuse = 13
From the figure,
Base side BC = 12
Hypotenuse AC = 13
Side AB is unknown here and it can be found by using Pythagoras theorem
Thus by applying Pythagoras theorem,
AC2 = AB2 + BC2
132 = AB2 + 122
Therefore,
AB2 = 132 – 122
AB2 = 169 – 144
AB2 = 25
AB = √25
AB = 5 …. (3)
Now, we know that
sin θ = Perpendicular side opposite to ∠θ / Hypotenuse
Thus, sin θ = AB/AC [from figure]
⇒ sin θ = 5/13… (4)
And, tan θ = sin θ / cos θ = (5/13) / (12/13)
⇒ tan θ = 12/13… (5)
Taking L.H.S we have
L.H.S = sin θ (1 – tan θ)
Substituting the value of sin θ and tan θ from equation (4) and (5)
We get,
15.
Solution:
Given, cot θ = 1/√3……. (1)
By definition we know that,
cot θ = 1/ tan θ
And, since tan θ = perpendicular side opposite to ∠θ / Base side adjacent to ∠θ
⇒ cot θ = Base side adjacent to ∠θ / perpendicular side opposite to ∠θ …… (2)[Since they are reciprocal to each other]
On comparing equation (1) and (2), we get
Base side adjacent to ∠θ = 1 and Perpendicular side opposite to ∠θ = √3
Therefore, the triangle formed is,
On substituting the values of known sides as AB = √3 and BC = 1
AC2 = (√3) + 1
AC2 = 3 + 1
AC2 = 4
AC = √4
Therefore, AC = 2 … (3)
Now, by definition
sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC
⇒ sin θ = √3/ 2 ……(4)
And, cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC
⇒ cos θ = 1/ 2 ….. (5)
Now, taking L.H.S we have
Substituting the values from equation (4) and (5), we have
16.
Solution:
Given, tan θ = 1/ √7 …..(1)
By definition, we know that
tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ ……(2)
On comparing equation (1) and (2), we have
Perpendicular side opposite to ∠θ = 1
Base side adjacent to ∠θ = √7
Thus, the triangle representing ∠ θ is,
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
By applying Pythagoras theorem, we have
AC2 = AB2 + BC2
AC2 = 12 + (√7)2
AC 2 = 1 + 7
AC2 = 8
AC = √8
⇒ AC = 2√2
By definition,
sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC
⇒ sin θ = 1/ 2√2
And, since cosec θ = 1/sin θ
⇒ cosec θ = 2√2 …….. (3)
Now,
cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC
⇒ cos θ = √7/ 2√2
And, since sec θ = 1/ sin θ
⇒ sec θ = 2√2/ √7 ……. (4)
Taking the L.H.S of the equation,
Substituting the value of cosec θ and sec θ from equation (3) and (4), we get
17. If sec θ = 5/4, find the value of
Solution:
Given,
sec θ = 5/4
We know that,
sec θ = 1/ cos θ
⇒ cos θ = 1/ (5/4) = 4/5 …… (1)
By definition,
cos θ = Base side adjacent to ∠θ / Hypotenuse …. (2)
On comparing equation (1) and (2), we have
Hypotenuse = 5
Base side adjacent to ∠θ = 4
Thus, the triangle representing ∠ θ is ABC.
Perpendicular side opposite to ∠θ, AB is unknown and it can be found by using Pythagoras theorem
By applying Pythagoras theorem, we have
AC2 = AB2 + BC2
AB2 = AC2 + BC2
AB2 = 52 – 42
AB2 = 25 – 16
AB = √9
⇒ AB = 3
By definition,
sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC
⇒ sin θ = 3/ 5 …..(3)
Now, tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ
⇒ tan θ = 3/ 4 ……(4)
And, since cot θ = 1/ tan θ
⇒ cot θ = 4/ 3 ……(5)
Now,
Substituting the value of sin θ, cos θ, cot θ and tan θ from the equations (1), (3), (4) and (5) we have,
= 12/7
Therefore,
18. If tan θ = 12/13, find the value of
Solution:
Given,
tan θ = 12/13 …….. (1)
We know that by definition,
tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ …… (2)
On comparing equation (1) and (2), we have
Perpendicular side opposite to ∠θ = 12
Base side adjacent to ∠θ = 13
Thus, in the triangle representing ∠ θ we have,
Hypotenuse AC is the unknown and it can be found by using Pythagoras theorem
So by applying Pythagoras theorem, we have
AC2 = 122 + 132
AC 2 = 144 + 169
AC2 = 313
⇒ AC = √313
By definition,
sin θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC
⇒ sin θ = 12/ √313…..(3)
And, cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC
⇒ cos θ = 13/ √313 …..(4)
Now, substituting the value of sin θ and cos θ from equation (3) and (4) respectively in the equation below
Therefore,
Exercise 5.2 Page No: 5.41
Evaluate each of the following:
1. sin 45∘ sin 30∘ + cos 45∘ cos 30∘
Solution:
2. sin 60∘ cos 30∘ + cos 60∘ sin 30∘
Solution:
3. cos 60∘ cos 45∘ – sin 60∘ sin 45∘
Solution:
4. sin2 30∘ + sin2 45∘ + sin2 60∘ + sin2 90∘
Solution:
5. cos2 30∘ + cos2 45∘ + cos2 60∘ + cos2 90∘
Solution:
6. tan2 30∘ + tan2 45∘ + tan2 60∘
Solution:
7. 2sin2 30∘ − 3cos2 45∘ + tan2 60∘
Solution:
8. sin2 30∘ cos245∘ + 4tan2 30∘ + (1/2) sin2 90∘ − 2cos2 90∘ + (1/24) cos20∘
Solution:
9. 4(sin4 60∘ + cos4 30∘) − 3(tan2 60∘ − tan2 45∘) + 5cos2 45∘
Solution:
10. (cosec2 45∘ sec2 30∘)(sin2 30∘ + 4cot2 45∘ − sec2 60∘)
Solution:
11. cosec3 30∘ cos60∘ tan3 45∘ sin2 90∘ sec2 45∘ cot30∘
Solution:
12. cot2 30∘ − 2cos2 60∘ − (3/4)sec2 45∘ – 4sec2 30∘
Solution:
Using trigonometric values, we have
13. (cos0∘ + sin45∘ + sin30∘)(sin90∘ − cos45∘ + cos60∘)
Solution:
(cos0∘ + sin45∘ + sin30∘)(sin90∘ − cos45∘ + cos60∘)
Using trigonometric values, we have
15. 4/cot2 30∘ + 1/sin2 60∘ − cos2 45∘
Solution:
16. 4(sin4 30∘ + cos2 60∘) − 3(cos2 45∘ − sin2 90∘) − sin2 60∘
Solution:
Using trigonometric values, we have
17.
Solution:
Using trigonometric values, we have
18.
Solution:
Using trigonometric values, we have
19.
Solution:
Using trigonometric values, we have
Find the value of x in each of the following: (20-25)
20. 2sin 3x = √3
Solution:
Given,
2 sin 3x = √3
sin 3x = √3/2
sin 3x = sin 60°
3x = 60°
x = 20°
Exercise 5.3 Page No: 5.52
1. Evalute the following:
(i) sin 20o/ cos 70o
(ii) cos 19o/ sin 71o
(iii) sin 21o/ cos 69o
(iv) tan 10o/ cot 80o
(v) sec 11o/ cosec 79o
Solution:
(i) We have,
sin 20o/ cos 70o = sin (90o – 70o)/ cos 70o = cos 70o/ cos70o = 1 [∵ sin (90 – θ) = cos θ]
(ii) We have,
cos 19o/ sin 71o = cos (90o – 71o)/ sin 71o = sin 71o/ sin 71o = 1 [∵ cos (90 – θ) = sin θ]
(iii) We have,
sin 21o/ cos 69o = sin (90o – 69o)/ cos 69o = cos 69o/ cos69o = 1 [∵ sin (90 – θ) = cos θ]
(iv) We have,
tan 10o/ cot 80o = tan (90o – 10o) / cot 80o = cot 80o/ cos80o = 1 [∵ tan (90 – θ) = cot θ]
(v) We have,
sec 11o/ cosec 79o = sec (90o – 79o)/ cosec 79o = cosec 79o/ cosec 79o = 1[∵ sec (90 – θ) = cosec θ]
2. Evaluate the following:
Solution:
We have, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= 12 + 12 = 1 + 1
= 2
(ii) cos 48°- sin 42°
Solution:
We know that, cos (90° − θ) = sin θ.
So,
cos 48° – sin 42° = cos (90° − 42°) – sin 42° = sin 42° – sin 42°= 0
Thus the value of cos 48° – sin 42° is 0.
Solution:
We have, [∵ cot (90 – θ) = tan θ and cos (90 – θ) = sin θ]
= 1 – 1/2(1)
= 1/2
Solution:
We have, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= 1 – 1
= 0
Solution:
We have, [∵ cot (90 – θ) = tan θ and tan (90 – θ) = cot θ]
= tan (90o – 35o)/ cot 55o + cot (90o – 12o)/ tan 12o – 1
= cot 55o/ cot 55o + tan 12o/ tan 12o – 1
= 1 + 1 – 1
= 1
Solution:
We have , [∵ sin (90 – θ) = cos θ and sec (90 – θ) = cosec θ]
= sec (90o – 20o)/ cosec 20o + sin (90o – 31o)/ cos 31o
= cosec 20o/ cosec 20o + cos 12o/ cos 12o
= 1 + 1
= 2
(vii) cosec 31° – sec 59°
Solution:
We have,
cosec 31° – sec 59°
Since, cosec (90 – θ) = cos θ
So,
cosec 31° – sec 59° = cosec (90° – 59o) – sec 59° = sec 59° – sec 59° = 0
Thus,
cosec 31° – sec 59° = 0
(viii) (sin 72° + cos 18°) (sin 72° – cos 18°)
Solution:
We know that,
sin (90 – θ) = cos θ
So, the given can be expressed as
(sin 72° + cos 18°) (sin (90 – 18)° – cos 18°)
= (sin 72° + cos 18°) (cos 18° – cos 18°)
= (sin 72° + cos 18°) x 0
= 0
(ix) sin 35° sin 55° – cos 35° cos 55°
Solution:
We know that,
sin (90 – θ) = cos θ
So, the given can be expressed as
sin (90 – 55)° sin (90 – 35)° – cos 35° cos 55°
= cos 55° cos 35° – cos 35° cos 55°
= 0
(x) tan 48° tan 23° tan 42° tan 67°
Solution:
We know that,
tan (90 – θ) = cot θ
So, the given can be expressed as
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 x 1 [∵ tan θ x cot θ = 1]
= 1
(xi) sec 50° sin 40° + cos 40° cosec 50°
Solution:
We know that,
sin (90 – θ) = cos θ and cos (90 – θ) = sin θ
So, the given can be expressed as
sec 50° sin (90 – 50)° + cos (90 – 50)° cosec 50°
= sec 50° cos 50° + sin 50° cosec 50°
= 1 + 1 [∵ sin θ x cosec θ = 1 and cos θ x sec θ = 1]
= 2
3. Express each one of the following in terms of trigonometric ratios of angles lying between 0o and 45o
(i) sin 59o + cos 56o (ii) tan 65o + cot 49o (iii) sec 76o + cosec 52o
(iv) cos 78o + sec 78o (v) cosec 54o + sin 72o (vi) cot 85o + cos 75o
(vii) sin 67o + cos 75o
Solution:
Using the below trigonometric ratios of complementary angles, we find the required
sin (90 – θ) = cos θ cosec (90 – θ) = sec θ
cos (90 – θ) = sin θ sec (90 – θ) = cosec θ
tan (90 – θ) = cot θ cot (90 – θ) = tan θ
(i) sin 59o + cos 56o = sin (90 – 31)o + cos (90 – 34)o = cos 31o + sin 34o
(ii) tan 65o + cot 49o = tan (90 – 25)o + cot (90 -41)o = cot 25o + tan 41o
(iii) sec 76o + cosec 52o = sec (90 – 14)o + cosec (90 – 38)o = cosec 14o + sec 38o
(iv) cos 78o + sec 78o = cos (90 – 12)o + sec (90 – 12)o = sin 12o + cosec 12o
(v) cosec 54o + sin 72o = cosec (90 – 36)o + sin (90 – 18)o = sec 36o + cos 18o
(vi) cot 85o + cos 75o = cot (90 – 5)o + cos (90 – 15)o = tan 5o + sin 15o
4. Express cos 75o + cot 75o in terms of angles between 0o and 30o.
Solution:
Given,
cos 75o + cot 75o
Since, cos (90 – θ) = sin θ and cot (90 – θ) = tan θ
cos 75o + cot 75o = cos (90 – 15)o + cot (90 – 15)o = sin 15o + tan 15o
Hence, cos 75o + cot 75o can be expressed as sin 15o + tan 15o
5. If sin 3A = cos (A – 26o), where 3A is an acute angle, find the value of A.
Solution:
Given,
sin 3A = cos (A – 26o)
Using cos (90 – θ) = sin θ, we have
sin 3A = sin (90o – (A – 26o))
Now, comparing both L.H.S and R.H.S
3A = 90o – (A – 26o)
3A + (A – 26o) = 90o
4A – 26o = 90o
4A = 116o
A = 116o/4
∴ A = 29o
6. If A, B, C are the interior angles of a triangle ABC, prove that
(i) tan ((C + A)/ 2) = cot (B/2) (ii) sin ((B + C)/ 2) = cos (A/2)
Solution:
We know that, in triangle ABC the sum of the angles i.e A + B + C = 180o
So, C + A = 180o – B ⇒ (C + A)/2 = 90o – B/2 …… (i)
And, B + C = 180o – A ⇒ (B + C)/2 = 90o – A/2 ……. (ii)
(i) L.H.S = tan ((C + A)/ 2)
⇒ tan ((C + A)/ 2) = tan (90o – B/2) [From (i)]
= cot (B/2) [∵ tan (90 – θ) = cot θ]
= R.H.S
- Hence Proved
(ii) L.H.S = sin ((B + C)/2)
⇒ sin ((B + C)/ 2) = sin (90o – A/2) [From (ii)]
= cos (A/2)
= R.H.S
- Hence Proved
7. Prove that:
(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1
(ii) sin 48° sec 48° + cos 48° cosec 42° = 2
Solution:
(i) Taking L.H.S = tan 20° tan 35° tan 45° tan 55° tan 70°
= tan (90° − 70°) tan (90° − 55°) tan 45°tan 55° tan70°
= cot 70°cot 55° tan 45° tan 55° tan 70° [∵ tan (90 – θ) = cot θ]
= (tan 70°cot 70°)(tan 55°cot 55°) tan 45° [∵ tan θ x cot θ = 1]
= 1 × 1 × 1 = 1
- Hence proved
(ii) Taking L.H.S = sin 48° sec 48° + cos 48° cosec 42°
= sin 48° sec (90° − 48°) + cos 48° cosec (90° − 48°)[∵sec (90 – θ) = cosec θ and cosec (90 – θ) = sec θ]
= sin 48°cosec 48° + cos 48°sec 48° [∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]
= 1 + 1 = 2
- Hence proved
(iii) Taking the L.H.S,
= 1 + 1 – 2
= 2 – 2
= 0
- Hence proved
(iv) Taking L.H.S,
= 1 + 1
= 2
- Hence proved
8. Prove the following:
(i) sinθ sin (90o – θ) – cos θ cos (90o – θ) = 0
Solution:
Taking the L.H.S,
sinθ sin (90o – θ) – cos θ cos (90o – θ)
= sin θ cos θ – cos θ sin θ [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= 0
(ii)
Solution:
Taking the L.H.S,
[∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]
= 1 + 1
= 2 = R.H.S
- Hence Proved
(iii)
Solution:
Taking the L.H.S, [∵ tan (90o – θ) = cot θ]
= 0 = R.H.S
- Hence Proved
(iv)
Solution:
Taking L.H.S, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= sin2 A = R.H.S
- Hence Proved
(v) sin (50o + θ) – cos (40o – θ) + tan 1o tan 10o tan 20o tan 70o tan 80o tan 89o = 1
Solution:
Taking the L.H.S,
= sin (50o + θ) – cos (40o – θ) + tan 1o tan 10o tan 20o tan 70o tan 80o tan 89o
= [sin (90o – (40o – θ))] – cos (40o – θ) + tan (90 – 89)o tan (90 – 80)o tan (90 – 70)o tan 70o tan 80o tan 89o [∵ sin (90 – θ) = cos θ]
= cos (40o – θ) – cos (40o – θ) + cot 89o cot 80o cot 70o tan 70o tan 80o tan 89o[∵ tan (90o – θ) = cot θ]
= 0 + (cot 89o x tan 89o) (cot 80o x tan 80o) (cot 70o x tan 70o)
= 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1]
= 1= R.H.S
- Hence Proved
RD Sharma Solutions for Class 10 Maths Chapter 5: Download PDF
RD Sharma Solutions for Class 10 Maths Chapter 5–Trigonometric Ratios
Download PDF: RD Sharma Solutions for Class 10 Maths Chapter 5–Trigonometric Ratios PDF
Chapterwise RD Sharma Solutions for Class 10 Maths :
- Chapter 1–Real Numbers
- Chapter 2–Polynomials
- Chapter 3–Pair of Linear Equations In Two Variables
- Chapter 4–Triangles
- Chapter 5–Trigonometric Ratios
- Chapter 6–Trigonometric Identities
- Chapter 7–Statistics
- Chapter 8–Quadratic Equations
- Chapter 9–Arithmetic Progressions
- Chapter 10–Circles
- Chapter 11–Constructions
- Chapter 12–Some Applications of Trigonometry
- Chapter 13–Probability
- Chapter 14–Co-ordinate Geometry
- Chapter 15–Areas Related To Circles
- Chapter 16–Surface Areas And Volumes
About RD Sharma
RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.
R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.