## RD Sharma Solutions Class 10 Maths Chapter 5 – Free PDF Download

**RD Sharma Solutions for Class 10 Maths Chapter 5 –** **Trigonometric Ratios** are provided here for students for easy access. A proper guidance tool in the right direction is essential for a student to excel in exams. The RD Sharma Solutions is one such powerful weapon prepared by our experts at BYJUâ€™S following the latest **CBSE** guidelines. This will definitely help the students secure high marks in their examinations as itâ€™s got well designed and detailed answers to problems in an easily understandable language.

The pdf of **RD Sharma Solutions Class 10 Chapter 5** Trigonometric Ratios are provided here. The questions present in this have been solved by BYJUâ€™S experts in Maths, and this will help students solve the problems without any difficulties.Â By practising sincerely, the students can obtain worthy results in board exams by using the RD Sharma Solutions for Class 10.

Trigonometry is the Science of measuring triangles. Further, in this chapter, you will learn about the trigonometric ratios and their relations between them. In addition, the **Trigonometric Ratios** of some specific angles are also discussed.

## Download the PDF of RD Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios here

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### RD Sharma Solutions for Class 10 Maths Chapter 5 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)Â sin A = 2/3**

**Solution: **

We have,

sin A = 2/3 â€¦â€¦..â€¦.. (1)

As we know, by sin definition;

sin AÂ =Â Â Perpendicular/ HypotenuseÂ = 2/3Â â€¦.(2)

By comparing eq. (1) and (2), weÂ have

Opposite side = 2 and Hypotenuse = 3

Now, on using Pythagoras theorem in Î”Â ABC

AC^{2Â }= AB^{2 +}Â BC^{2}

Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get

â‡’ 3^{2}Â = AB^{2}Â + 2^{2}

AB^{2Â }= 3^{2}Â â€“ 2^{2}

AB^{2}Â = 9 â€“ 4

AB^{2 =Â }5

AB = âˆš5

Hence, Base =Â âˆš5

By definition,

cos A = Base/Hypotenuse

â‡’ cos A = âˆš5/3

Since, cosec A =Â 1/sin A = Hypotenuse/Perpendicular

â‡’ cosec A = 3/2

And, sec A =Â Hypotenuse/Base

â‡’ sec A =Â 3/âˆš5

And, tan A =Â Perpendicular/Base

â‡’ tan A =Â Â 2/âˆš5

And, cot A =Â 1/ tan A = Base/Perpendicular

â‡’ cot A =Â âˆš5/2

**(ii) cos A = 4/5**

**Solution: **

We have,

cos A = 4/5Â â€¦â€¦.â€¦. (1)

As we know, by cos defination

cos A = Base/HypotenuseÂ â€¦. (2)

By comparing eq. (1) and (2), weÂ get

Base = 4 and Hypotenuse = 5

Now, using Pythagoras theorem in Î”Â ABC

AC^{2Â }= AB^{2Â }+ BC^{2}

Putting the value of base (AB) and hypotenuse (AC) and for the perpendicular (BC), we get

5^{2}Â = 4^{2 }+ BC^{2}

BC^{2}Â = 5^{2}Â â€“ 4^{2}

BC^{2 }= 25 â€“ 16

BC^{2}Â = 9

BC= 3

Hence, Perpendicular = 3

By definition,

sin AÂ =Â Perpendicular/Hypotenuse

â‡’ sin A = 3/5

Then, cosec A =Â 1/sin A

â‡’ cosec A=Â 1/ (3/5) = 5/3 = Hypotenuse/Perependicular

And, sec A = 1/cos A

â‡’ sec A =Hypotenuse/Base

sec A =Â 5/4

And, tan A =Â Perpendicular/Base

â‡’ tan A = 3/4

Next, cot A =Â 1/tan A = Base/Perpendicular

âˆ´ cot A =Â 4/3

**(iii) tan Î¸ = 11/1**

**Solution: **

We have, tanÂ Î¸ = 11â€¦..â€¦. (1)

By definition,

tanÂ Î¸ = Perpendicular/ Baseâ€¦. (2)

On Comparing eq. (1) and (2), weÂ get;

Base = 1 andÂ Perpendicular = 5

Now, using Pythagoras theorem in Î”Â ABC.

AC^{2}Â = AB^{2}Â + BC^{2}

Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get;

AC^{2}Â = 1^{2}Â + 11^{2}

AC^{2}Â = 1 + 121

AC^{2}= 122

AC= âˆš122

Hence, hypotenuse = âˆš122

By definition,

sin = Perpendicular/Hypotenuse

â‡’ sinÂ Î¸ = 11/âˆš122

And, cosecÂ Î¸Â = 1/sinÂ Î¸

â‡’ cosecÂ Î¸Â = âˆš122/11

Next, cosÂ Î¸ = Base/ Hypotenuse

â‡’ cosÂ Î¸ = 1/âˆš122

And, secÂ Î¸ = 1/cosÂ Î¸

â‡’ secÂ Î¸ = âˆš122/1 =Â âˆš122

And, cotÂ Î¸ Â = 1/tanÂ Î¸

âˆ´Â cotÂ Î¸ = 1/11

**(iv) sin Î¸ = 11/15**

**Solution: **

We have, Â sin Î¸ = 11/15Â â€¦â€¦â€¦. (1)

By definition,

sin Î¸ = Perpendicular/ HypotenuseÂ â€¦. (2)

On Comparing eq. (1) and (2), weÂ get;

Perpendicular = 11 and Hypotenuse= 15

Now, using Pythagoras theorem in Î”Â ABC

AC^{2}Â = AB^{2}Â + BC^{2}

Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have

15^{2}Â = AB^{2}Â +11^{2}

AB^{2}Â = 15^{2Â }â€“ 11^{2}

AB^{2 }= 225 â€“ 121

AB^{2Â }= 104

AB =Â âˆš104

AB=Â âˆš (2Ã—2Ã—2Ã—13)

AB= 2âˆš(2Ã—13)

AB= 2âˆš26

Hence, Base = 2âˆš26

By definition,

cosÂ Î¸ = Base/Hypotenuse

âˆ´Â cosÎ¸ = 2âˆš26/ 15

And, cosecÂ Î¸Â = 1/sinÂ Î¸

âˆ´ cosecÂ Î¸Â = 15/11

And, secÎ¸Â = Hypotenuse/Base

âˆ´ secÎ¸ =15/ 2âˆš26

And,Â tan Î¸ = Perpendicular/Base

âˆ´Â tanÎ¸ =11/ 2âˆš26

And,Â cot Î¸ = Base/Perpendicular

âˆ´Â cotÎ¸ =2âˆš26/ 11

**Â (v) tan Î± = 5/12**

**Solution:**

We have, Â tanÂ Î± = 5/12Â â€¦. (1)

By definition,

tan Î± = Perpendicular/Baseâ€¦. (2)

On Comparing eq. (1) and (2), weÂ get

Base = 12 and Perpendicular side = 5

Now, using Pythagoras theorem in Î”Â ABC

AC^{2}Â = AB^{2}Â + BC^{2}

Putting the value of base (AB) and the perpendicular (BC) to get hypotenuse (AC), we have

AC^{2}Â = 12^{2}Â + 5^{2}

AC^{2}Â = 144 + 25

AC^{2}= 169

AC = 13 [After taking sq root on both sides]

Hence, Hypotenuse = 13

By definition,

sinÂ Î±Â Â = Perpendicular/Hypotenuse

âˆ´Â sinÂ Î± = 5/13

And, cosecÂ Î±Â Â = Hypotenuse/Perpendicular

âˆ´ cosecÂ Î±Â = 13/5

And,Â cosÂ Î± = Base/Hypotenuse

âˆ´Â cosÂ Î± = 12/13

And,Â secÂ Î± =1/cosÂ Î±

âˆ´ secÂ Î± = 13/12

And, tanÂ Î± = sinÂ Î±/cosÂ Î±

âˆ´ tanÂ Î±=5/12

Since, cotÂ Î± = 1/tanÂ Î±

âˆ´Â cotÂ Î± =12/5

**Â (vi) sin Î¸ = âˆš3/2**

**Solution: **

We have,Â sinÂ Î¸ =Â âˆš3/2Â â€¦â€¦â€¦â€¦. (1)

By definition,

sinÂ Î¸ = Perpendicular/ Hypotenuseâ€¦.(2)

On Comparing eq. (1) and (2), weÂ get;

Perpendicular =Â âˆš3 andÂ Hypotenuse = 2

Now, using Pythagoras theorem in Î”Â ABC

AC^{2}Â = AB^{2}Â + BC^{2}

Putting the value of perpendicular (BC) and hypotenuse (AC) and get the base (AB), we get;

2^{2}Â = AB^{2}Â + (âˆš3)^{2}

AB^{2}Â = 2^{2}Â â€“ (âˆš3)^{2}

AB^{2}Â = 4 â€“ 3

AB^{2}Â = 1

AB = 1

Thus, Base = 1

By definition,

cosÂ Î¸ = Base/Hypotenuse

âˆ´Â cos Î¸ = 1/2

And, cosecÂ Î¸Â = 1/sinÂ Î¸

Or cosecÂ Î¸= Hypotenuse/Perpendicualar

âˆ´ cosecÂ Î¸Â =2/âˆš3

And,Â secÂ Î¸ = Hypotenuse/Base

âˆ´Â sec Î¸ = 2/1

And,Â tanÂ Î¸ = Perpendicula/Base

âˆ´Â tanÂ Î¸ = âˆš3/1

And,Â cotÂ Î¸ = Base/Perpendicular

âˆ´Â cotÂ Î¸ = 1/âˆš3

**(vii) cos Î¸ = 7/25**

**Solution: **

We have,Â cos Î¸ = 7/25Â â€¦â€¦â€¦.. (1)

By definition,

cosÂ Î¸ = Base/Hypotenuse

On Comparing eq. (1) and (2), weÂ get;

Base = 7 and Hypotenuse = 25

Now, using Pythagoras theorem in Î”Â ABC

AC^{2}= AB^{2}Â + BC^{2}

Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC)

25^{2}Â = 7^{2Â }+BC^{2}

BC^{2}Â = 25^{2}Â â€“ 7^{2}

BC^{2}Â = 625 â€“ 49

BC^{2} = 576

BC=Â âˆš576

BC= 24

Hence, Perpendicular side = 24

By definition,

sin Î¸ = perpendicular/Hypotenuse

âˆ´ sin Î¸Â = 24/25

Since, cosecÂ Î¸Â = 1/sinÂ Î¸

Also, cosecÂ Î¸= Hypotenuse/Perpendicualar

âˆ´ cosecÂ Î¸Â = 25/24

Since,Â secÂ Î¸ = 1/cosecÂ Î¸

Also,Â sec Î¸ = Hypotenuse/Base

âˆ´ secÂ Î¸ = 25/7

Since,Â tan Î¸ = Perpendicular/Base

âˆ´Â tanÂ Î¸ = 24/7

Now,Â cot = 1/tanÂ Î¸

So,Â cot Î¸ = Base/Perpendicular

âˆ´ cotÂ Î¸ = 7/24

**(viii) tan Î¸ = 8/15**

**Solution:**

We have,Â tanÂ Î¸ = 8/15Â â€¦â€¦â€¦â€¦. (1)

By definition,

tan Î¸ = Perpendicular/BaseÂ â€¦. (2)

On Comparing eq. (1) and (2), weÂ get;

Base = 15 and Perpendicular = 8

Now, using Pythagoras theorem in Î”Â ABC

AC^{2 }= 15^{2}Â + 8^{2}

AC^{2 }= 225 + 64

AC^{2}Â = 289

AC =Â âˆš289

AC = 17

Hence, Hypotenuse = 17

By definition,

Since,Â sin Î¸ = perpendicular/Hypotenuse

âˆ´Â sin Î¸ = 8/17

Since, cosecÂ Î¸Â = 1/sin Î¸

Also, cosecÂ Î¸ = Hypotenuse/Perpendicualar

âˆ´ cosec Î¸ = 17/8

Since,Â cosÂ Î¸ = Base/Hypotenuse

âˆ´Â cos Î¸ = 15/17

Since,Â sec Î¸ = 1/cos Î¸

Also,Â secÂ Î¸ = Hypotenuse/Base

âˆ´ sec Î¸ = 17/15

Since, cot Î¸ = 1/tan Î¸

Also, Â cot Î¸ = Base/Perpendicular

âˆ´ cot Î¸ = 15/8

**(ix) cot Î¸ = 12/5**

**Solution: **

We have, cotÂ Î¸ = 12/5 â€¦â€¦â€¦â€¦. (1)

By definition,

cot Î¸ = 1/tan Î¸

cot Î¸ = Base/PerpendicularÂ â€¦â€¦. (2)

On Comparing eq. (1) and (2), weÂ have

Base = 12 and Perpendicular side = 5

Now, using Pythagoras theorem in Î”Â ABC

AC^{2}= AB^{2}Â + BC^{2}

Putting the value of base (AB) and perpendicular (BC) to get the hypotenuse (AC);

AC^{2}Â = 12^{2}Â + 5^{2}

AC^{2}= 144 + 25

AC^{2}Â = 169

AC =Â âˆš169

AC = 13

Hence, Hypotenuse = 13

By definition,

Since,Â sin Î¸ = perpendicular/Hypotenuse

âˆ´Â sin Î¸= 5/13

Since, cosecÂ Î¸Â = 1/sin Î¸

Also, cosecÂ Î¸= Hypotenuse/Perpendicualar

âˆ´ cosecÂ Î¸Â = 13/5

Since,Â cosÂ Î¸ = Base/Hypotenuse

âˆ´Â cos Î¸ = 12/13

Since,Â sec Î¸ = 1/cosÎ¸

Also,Â secÂ Î¸ = Hypotenuse/Base

âˆ´ sec Î¸ = 13/12

Since,Â tanÎ¸ = 1/cot Î¸

Also,Â tan Î¸ = Perpendicular/Base

âˆ´ tan Î¸ = 5/12

**(x) Â sec Î¸ = 13/5**

**Solution: **

We have, sec Î¸ = 13/5â€¦â€¦.â€¦ (1)

By definition,

secÂ Î¸ = Hypotenuse/Baseâ€¦â€¦â€¦â€¦. (2)

On Comparing eq. (1) and (2), weÂ get

Base = 5 andÂ Hypotenuse = 13

Now, using Pythagoras theorem in Î”Â ABC

AC^{2Â }= AB^{2}Â + BC^{2}

And. putting the value of base side (AB) and hypotenuse (AC) to get the perpendicular side (BC)

13^{2}Â = 5^{2}Â + BC^{2}

BC^{2}Â = 13^{2Â }â€“ 5^{2}

BC^{2}=169 â€“ 25

BC^{2}= 144

BC=Â âˆš144

BC = 12

Hence, Perpendicular = 12

By definition,

Since, Â sin Î¸ = perpendicular/Hypotenuse

âˆ´ sin Î¸= 12/13

Since, cosecÂ Î¸= 1/ sin Î¸

Also, cosecÂ Î¸= Hypotenuse/Perpendicualar

âˆ´Â cosecÂ Î¸Â = 13/12

Since,Â cos Î¸= 1/sec Î¸

Also,Â cosÂ Î¸ = Base/Hypotenuse

âˆ´ cos Î¸ = 5/13

Since,Â tan Î¸ = Perpendicular/Base

âˆ´Â tan Î¸ = 12/5

Since,Â cot Î¸ = 1/tan Î¸

Also,Â cot Î¸ = Base/Perpendicular

âˆ´ cot Î¸ = 5/12

**(xi) Â cosec Î¸ = âˆš10**

**Solution: **

We have, cosecÂ Î¸Â = âˆš10/1Â Â â€¦â€¦..â€¦ (1)

By definition,

cosecÂ Î¸ = Hypotenuse/ Perpendicualar â€¦â€¦.â€¦.(2)

And, cosecÎ¸ = 1/sin Î¸

On comparing eq.(1) and(2), we get

Perpendicular side = 1 andÂ Hypotenuse =Â âˆš10

Now, using Pythagoras theorem in Î”Â ABC

AC^{2Â }= AB^{2}Â + BC^{2}

Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base side (AB)

(âˆš10)^{2}Â = AB^{2}Â + 1^{2}

AB^{2}= (âˆš10)^{2}Â â€“ 1^{2}

AB^{2}= 10 â€“ 1

AB =Â âˆš9

AB = 3

So, Base side = 3

By definition,

Since,Â sin Î¸ = Perpendicular/Hypotenuse

âˆ´Â sin Î¸ = 1/âˆš10

Since,Â cosÂ Î¸ = Base/Hypotenuse

âˆ´Â cos Î¸ = 3/âˆš10

Since,Â sec Î¸ = 1/cos Î¸

Also, secÂ Î¸ = Hypotenuse/Base

âˆ´Â sec Î¸ = âˆš10/3

Since,Â tan Î¸ = Perpendicular/Base

âˆ´Â tan Î¸ = 1/3

Since,Â cot Î¸ = 1/tan Î¸

âˆ´ cot Î¸ = 3/1

**(xii)Â cos Î¸ =12/15**

**Solution: **

We have;Â cosÂ Î¸ = 12/15Â â€¦â€¦â€¦. (1)

By definition,

cosÂ Î¸ = Base/Hypotenuseâ€¦â€¦â€¦ (2)

By comparing eq. (1) and (2), weÂ get;

Base =12 and Hypotenuse = 15

Now, using Pythagoras theorem in Î”Â ABC, we get

AC^{2}Â = AB^{2}+ BC^{2}

Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC);

15^{2}Â = 12^{2}Â + BC^{2}

BC^{2}Â = 15^{2}Â â€“ 12^{2}

BC^{2}Â = 225 â€“ 144

BCÂ ^{2}= 81

BC =Â âˆš81

BC = 9

So, Perpendicular = 9

By definition,

Since,Â sin Î¸ = perpendicular/Hypotenuse

âˆ´Â sinÂ Î¸ = 9/15 = 3/5

Since, cosecÂ Î¸Â = 1/sinÂ Î¸

Also, cosecÂ Î¸ = Hypotenuse/Perpendicualar

âˆ´Â cosecÂ Î¸= 15/9 = 5/3

Since,Â sec Î¸ = 1/cos Î¸

Also,Â secÂ Î¸ = Hypotenuse/Base

âˆ´ secÂ Î¸ = 15/12 = 5/4

Since,Â tan Î¸ = Perpendicular/Base

âˆ´Â tanÂ Î¸ = 9/12 = 3/4

Since,Â cotÂ Î¸ = 1/tanÂ Î¸

Also,Â cot Î¸ = Base/Perpendicular

âˆ´ cotÂ Î¸ = 12/9 = 4/3

**2. In a â–³ 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 â–³ABC, AB = 24 cm, BC = 7cm and âˆ ABCÂ = 90^{o}

To find: sin A, cos A

By using Pythagoras theorem in â–³ABC we have

AC^{2}Â = AB^{2}Â + BC^{2}

AC^{2}Â = 24^{2}Â + 7^{2}

AC^{2}Â = 576 + 49

AC^{2}= 625

AC =Â âˆš625

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

**Â **

**(ii) **Given:Â In â–³ABC , AB = 24 cm and BC = 7cm and âˆ ABCÂ = 90^{o}

To find: sin C, cos C

By using Pythagoras theorem in â–³ABC we have

AC^{2}Â = AB^{2}Â + BC^{2}

AC^{2}Â = 24^{2}Â + 7^{2}

AC^{2}Â = 576 + 49

AC^{2}= 625

AC =Â âˆš625

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?Â **

**Solution: **

By using Pythagoras theorem in â–³PQR, we have

PR^{2}Â = PQ^{2}Â + QR^{2}

Putting the length of given side PR and PQ in the above equation

13^{2 }= 12^{2}Â + QR^{2}

QR^{2}Â = 13^{2}Â â€“ 12^{2}

QR^{2}Â = 169 â€“ 144

QR^{2 }= 25

QR =Â âˆš25 = 5

By definition,

tan P = Perpendicular side opposite to P/ Base side adjacent to angle P

tan P = QR/PQ

tan P = 5/12Â â€¦â€¦â€¦. (1)

And,

cot R= Base/Perpendicular

cot R= QR/PQ

cot R= 5/12Â â€¦. (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.

âˆ´ tan P = cot R

**Yes, tan P = cot R =Â 5/12**

**4. If sin A =Â 9/41, compute cos A and tan A.**

**Solution:**

** **

Given that,Â sin A = 9/41Â â€¦â€¦â€¦â€¦. (1)

Required to find: cos A, tan A

By definition, we know that

sin A = Perpendicular/ Hypotenuseâ€¦â€¦â€¦â€¦â€¦(2)

On Comparing eq. (1) and (2), weÂ get;

Perpendicular side = 9 and Hypotenuse = 41

Letâ€™s construct â–³ABCÂ as shown below,

And, here the length of base AB is unknown.

Thus, by using Pythagoras theorem in â–³ABC, we get;

AC^{2}Â = AB^{2}Â + BC^{2}

41^{2}Â = AB^{2}Â + 9^{2}

AB^{2}Â = 41^{2}Â â€“ 9^{2}

AB^{2}Â = 168 â€“ 81

AB= 1600

AB =Â âˆš1600

AB = 40

â‡’ 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.Â 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

â‡’ cot A = 8/15 â€¦â€¦.(1)

And we know,

cot A = 1/tan A

Also by definition,

Cot A = Base side adjacent to âˆ A/ Perpendicular side opposite to âˆ AÂ â€¦. (2)

On comparing equation (1) and (2), we get;

Base side adjacent to âˆ AÂ = 8

Perpendicular side opposite to âˆ AÂ = 15

So, by using Pythagoras theorem to â–³ABC, we have

AC^{2}Â = AB^{2}Â +BC^{2}

Substituting values for sides from the figure

AC^{2}Â = 8^{2}Â + 15^{2}

AC^{2}Â = 64 + 225

AC^{2}Â = 289

AC =Â âˆš289

AC = 17

Therefore, hypotenuse =17

By definition,

sin A = Perpendicular/Hypotenuse

â‡’ sin A= BC/AC

sin A= 15/17 (using values from the above)

Also,

sec A= 1/ cos A

â‡’ secA = Hypotenuse/ Base side adjacent to âˆ A

âˆ´ sec A= 17/8

**Â 6. In â–³PQR, right-angled at Q, PQ = 4cm and RQ = 3 cm. Find the value of sin P,Â sin R, sec P and sec R.**

**Solution: **

** **

Given:

â–³PQRÂ is right-angled at Q.

PQ = 4cm

RQ = 3cm

Required to find: sin P, sin R, sec P, sec R

Given â–³PQR,

By using Pythagoras theorem to â–³PQR, we get

PR^{2}Â = PQ^{2}Â +RQ^{2}

Substituting the respective values,

PR^{2}Â = 4^{2}Â +3^{2}

PR^{2}Â = 16 + 9

PR^{2}Â = 25

PR =Â âˆš25

PR = 5

â‡’ Hypotenuse =5

By definition,

sin P = Perpendicular side opposite to angle P/ Hypotenuse

sin P = RQ/ PR

â‡’ sin P = 3/5

And,

sin R = Perpendicular side opposite to angle R/ Hypotenuse

sin R = PQ/ PR

â‡’ sin R = 4/5

And,

sec P=1/cos P

secP = Hypotenuse/ Base side adjacent to âˆ P

sec P = PR/ PQ

â‡’ sec P = 5/4

Now,

sec R = 1/cos R

secR = Hypotenuse/ Base side adjacent to âˆ R

sec R = PR/ RQ

â‡’ sec R = 5/3

**7.** **IfÂ cot Î¸ = 7/8, evaluate**

**Â Â Â Â (i)Â Â Â (1+sin Î¸)(1â€“sin Î¸)/ (1+cos Î¸)(1â€“cos Î¸)**

**Â Â Â Â (ii) Â cot ^{2 }Î¸**

**Solution: **

**(i) **Required to evaluate:

, given = cot Î¸ = 7/8

Taking the numerator, we have

(1+sin Î¸)(1â€“sin Î¸) = 1 â€“ sin^{2} Î¸ [Since, (a+b)(a-b) = a^{2} â€“ b^{2}]

Similarly,

(1+cos Î¸)(1â€“cos Î¸) = 1 â€“ cos^{2} Î¸

We know that,

sin^{2} Î¸ + cos^{2} Î¸ = 1

â‡’ 1 â€“ cos^{2} Î¸ = sin^{2} Î¸

And,

1 â€“ sin^{2} Î¸ = cos^{2} Î¸

Thus,

(1+sin Î¸)(1 â€“sin Î¸) = 1 â€“ sin^{2} Î¸ = cos^{2} Î¸

(1+cos Î¸)(1â€“cos Î¸) = 1 â€“ cos^{2} Î¸ = sin^{2} Î¸

â‡’

= cos^{2} Î¸/ sin^{2} Î¸

= (cos Î¸/sin Î¸)^{2}

And, we know that (cos Î¸/sin Î¸) = cot Î¸

â‡’

**= **(cot Î¸)^{2}

= (7/8)^{2}

= 49/ 64

**(ii) **Given,

cot Î¸ = 7/8

So, by squaring on both sides we get

(cot Î¸)^{2} = (7/8)^{2}

âˆ´ cot Î¸^{2} = 49/64

**8. IfÂ 3cot A = 4, check whetherÂ (1â€“tan ^{2}A)/(1+tan^{2}A) = (cos^{2}A â€“ sin^{2}A)Â or not.**

**Solution:**

** **

Given,

3cot A = 4

â‡’ cot A = 4/3

By definition,

tan A = 1/ Cot A = 1/ (4/3)

â‡’ tan A = 3/4

Thus,

Base side adjacent to âˆ A = 4

Perpendicular side opposite to âˆ AÂ = 3

InÂ Î”ABC, Hypotenuse is unknown

Thus, by applying Pythagoras theorem in Î”ABC

We get

AC^{2Â }= AB^{2}Â + BC^{2}

AC^{2}Â = 4^{2}Â + 3^{2}

AC^{2}Â = 16 + 9

AC^{2}Â = 25

AC = âˆš25

AC = 5

Hence, hypotenuse = 5

Now, we can find that

sin A = opposite side to âˆ A/ Hypotenuse = 3/5

And,

cos A = adjacent side to âˆ A/ Hypotenuse = 4/5

Taking the LHS,

Thus, LHS = 7/25

Now, taking RHS

**9. If tan Î¸ = a/b, find the value of (cos Î¸ + sin Î¸)/ (cos Î¸ – sin Î¸) **

**Solution: **

Given,

tan Î¸ = a/b

And, we know by definition that

tan Î¸ = 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

Hypotenuse^{2} = opposite side^{2} + adjacent side^{2}

â‡’ Hypotenuse = âˆš(a^{2} + b^{2})

So, by definition

sin Î¸ = opposite side/ Hypotenuse

sin Î¸ = a/ âˆš(a^{2} + b^{2})

And,

cos Î¸ = adjacent side/ Hypotenuse

cos Î¸ = b/ âˆš(a^{2} + b^{2})

Now,

After substituting for cos Î¸ and sin Î¸, we have

âˆ´

Hence Proved.

**10. IfÂ 3 tan Î¸ = 4, find the value of**

**Solution: **

Given, 3 tan Î¸ = 4

â‡’ tan Î¸ = 4/3

From, letâ€™s divide the numerator and denominator by cos Î¸.

We get,

(4 â€“ tan Î¸) / (2 + tan Î¸)

â‡’ (4 â€“ (4/3)) / (2 + (4/3)) [using the value of tan Î¸]

â‡’ (12 â€“ 4) / (6 + 4) [After taking LCM and cancelling it]

â‡’ 8/10 = 4/5

âˆ´ = 4/5

**11. IfÂ 3 cot Î¸ = 2, find the value of **

**Solution: **

Given, 3 cot Î¸ = 2

â‡’ cot Î¸ = 2/3

From, letâ€™s divide the numerator and denominator by sin Î¸.

We get,

(4 â€“3 cot Î¸) / (2 + 6 cot Î¸)

â‡’ (4 â€“ 3(2/3)) / (2 + 6(2/3)) [using the value of tan Î¸]

â‡’ (4 â€“ 2) / (2 + 4) [After taking LCM and simplifying it]

â‡’ 2/6 = 1/3

âˆ´ = 1/3

**12. IfÂ tan Î¸ = a/b, prove that**

**Solution: **

Given, tan Î¸ = a/b

From LHS, letâ€™s divide the numerator and denominator by cos Î¸.

And we get,

(a tan Î¸ â€“ b) / (a tan Î¸ + b)

â‡’ (a(a/b) â€“ b) / (a(a/b) + b) [using the value of tan Î¸]

â‡’ (a^{2} â€“ b^{2})/b^{2} / (a^{2} + b^{2})/b^{2} [After taking LCM and simplifying it]

â‡’ (a^{2} â€“ b^{2})/ (a^{2} + b^{2})

= RHS

– Hence Proved

**13. IfÂ sec Î¸ = 13/5, show that**

**Solution: **

** **

Given,

sec Î¸ = 13/5

We know that,

sec Î¸ = 1/ cos Î¸

â‡’ cos Î¸ = 1/ sec Î¸ = 1/ (13/5)

âˆ´ cos Î¸ = 5/13 â€¦â€¦. (1)

By definition,

cos Î¸ = adjacent side/ hypotenuse â€¦.. (2)

Comparing (1) and (2), we have

Adjacent side = 5 and hypotenuse = 13

By Pythagoras theorem,

Opposite side = âˆš((hypotenuse)^{ 2} â€“ (adjacent side)^{2})

= âˆš(13^{2} – 5^{2})

= âˆš(169 â€“ 25)

= âˆš(144)

= 12

Thus, opposite side = 12

By definition,

tan Î¸ = opposite side/ adjacent side

âˆ´ tan Î¸ = 12/ 5

From, letâ€™s divide the numerator and denominator by cos Î¸.

We get,

(2 tan Î¸ â€“ 3) / (4 tan Î¸ â€“ 9)

â‡’ (2(12/5) â€“ 3) / (4(12/5) â€“ 9) [using the value of tan Î¸]

â‡’ (24 â€“ 15) / (48 – 45) [After taking LCM and cancelling it]

â‡’ 9/3 = 3

âˆ´ = 3

**14. IfÂ cos Î¸ = 12/13, show thatÂ sin Î¸(1 â€“ tan Î¸) = 35/156**

**Solution:**

Given, cos Î¸ = 12/13â€¦â€¦ (1)

By definition we know that,

cos Î¸ = Base side adjacent to âˆ Î¸ / Hypotenuseâ€¦â€¦. (2)

When comparing equation (1) and (2), we get

Base side adjacent toÂ âˆ Î¸Â = 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,

AC^{2 }= AB^{2}Â + BC^{2}

13^{2 }= AB^{2}Â + 12^{2}

Therefore,

AB^{2 }= 13^{2Â }â€“ 12^{2}

AB^{2 }= 169 â€“ 144

AB^{2} = 25

AB =Â âˆš25

AB = 5 â€¦. (3)

Now, we know that

sin Î¸ = Perpendicular side opposite to âˆ Î¸ / Hypotenuse

Thus, sin Î¸ = AB/AC [from figure]

â‡’ sin Î¸ = 5/13â€¦ (4)

And, tan Î¸ = sin Î¸ / cos Î¸ = (5/13) / (12/13)

â‡’ tan Î¸ = 12/13â€¦ (5)

Taking L.H.S we have

L.H.S =Â sin Î¸ (1 â€“ tan Î¸)

Substituting the value ofÂ sin Î¸Â andÂ tan Î¸ from equation (4) and (5)

We get,

**15. **

**Solution: **

Given, cot Î¸ = 1/**âˆš**3……. (1)

By definition we know that,

cot Î¸ = 1/ tan Î¸

And, since tan Î¸ = perpendicular side opposite to âˆ Î¸Â / Base side adjacent to âˆ Î¸

â‡’ cot Î¸ = Base side adjacent to âˆ Î¸Â / perpendicular side opposite to âˆ Î¸Â â€¦â€¦ (2)

[Since they are reciprocal to each other]On comparing equation (1) and (2), we get

Base side adjacent toÂ âˆ Î¸Â = 1 and Perpendicular side opposite toÂ âˆ Î¸ = âˆš3

Therefore, the triangle formed is,

On substituting the values of known sides as AB = âˆš3 and BC = 1

AC^{2 }= (âˆš3) + 1

AC^{2}Â = 3 + 1

AC^{2Â }= 4

AC = âˆš4

Therefore, AC = 2 â€¦ Â Â (3)

Now, by definition

sin Î¸ = Perpendicular side opposite to âˆ Î¸ / Hypotenuse = AB / AC

â‡’ sin Î¸ = âˆš3/ 2 â€¦â€¦(4)

And, cos Î¸ = Base side adjacent to âˆ Î¸ / Hypotenuse = BC / AC

â‡’ cos Î¸ = 1/ 2 â€¦.. (5)

Now, taking L.H.S we have

Substituting the values from equation (4) and (5), we have

**16.**

**Solution:**

Given, tan Î¸ = 1/ âˆš7 â€¦..(1)

By definition, we know that

tan Î¸ = Perpendicular side opposite to âˆ Î¸ / Base side adjacent to âˆ Î¸ â€¦â€¦(2)

On comparing equation (1) and (2), we have

Perpendicular side opposite toÂ âˆ Î¸Â = 1

Base side adjacent toÂ âˆ Î¸ = âˆš7

Thus, the triangle representingÂ âˆ Î¸Â is,

Hypotenuse AC is unknown and it can be found by using Pythagoras theorem

By applying Pythagoras theorem, we have

AC^{2Â }= AB^{2}Â + BC^{2}

AC^{2Â }= 1^{2}Â + (âˆš7)^{2}

ACÂ ^{2}Â = 1 + 7

AC^{2}Â = 8

AC = âˆš8

â‡’ AC = 2âˆš2

By definition,

sin Î¸ = Perpendicular side opposite to âˆ Î¸ / Hypotenuse = AB / AC

â‡’ sin Î¸ = 1/ 2âˆš2

And, since cosec Î¸ = 1/sin Î¸

â‡’ cosec Î¸ = 2âˆš2 â€¦â€¦.. (3)

Now,

cos Î¸ = Base side adjacent to âˆ Î¸ / Hypotenuse = BC / AC

â‡’ cos Î¸ = âˆš7/ 2âˆš2

And, since sec Î¸ = 1/ sin Î¸

â‡’ sec Î¸ = 2âˆš2/ âˆš7 â€¦â€¦. (4)

Taking the L.H.S of the equation,

Substituting the value of cosec Î¸Â and sec Î¸Â from equation (3) and (4), we get

**17. IfÂ sec Î¸ = 5/4, find the value of**

**Solution:**

Given,

sec Î¸ = 5/4

We know that,

sec Î¸ = 1/ cos Î¸

â‡’ cos Î¸ = 1/ (5/4) = 4/5 â€¦â€¦ (1)

By definition,

cos Î¸ = Base side adjacent to âˆ Î¸ / Hypotenuse â€¦. (2)

On comparing equation (1) and (2), we have

HypotenuseÂ = 5

Base side adjacent toÂ âˆ Î¸ = 4

Thus, the triangle representingÂ âˆ Î¸Â is ABC.

Perpendicular side opposite to âˆ Î¸, AB is unknown and it can be found by using Pythagoras theorem

By applying Pythagoras theorem, we have

AC^{2Â }= AB^{2}Â + BC^{2}

AB^{2Â }= AC^{2}Â + BC^{2}

AB^{2}Â = 5^{2} â€“ 4^{2}

AB^{2}Â = 25 â€“ 16

AB = âˆš9

â‡’ AB = 3

By definition,

sin Î¸ = Perpendicular side opposite to âˆ Î¸ / Hypotenuse = AB / AC

â‡’ sin Î¸ = 3/ 5 â€¦..(3)

Now, tan Î¸ = Perpendicular side opposite to âˆ Î¸ / Base side adjacent to âˆ Î¸

â‡’ tan Î¸ = 3/ 4 â€¦â€¦(4)

And, since cot Î¸ = 1/ tan Î¸

â‡’ cot Î¸ = 4/ 3 â€¦â€¦(5)

Now,

Substituting the value ofÂ sin Î¸, cos Î¸,Â cot Î¸Â andÂ tan Î¸Â from the equations (1), (3), (4) and (5) we have,

= 12/7

Therefore,

**18. IfÂ tan Î¸ = 12/13, find the value of**

**Solution:**

Given,

tan Î¸ = 12/13 â€¦â€¦.. (1)

We know that by definition,

tan Î¸ = Perpendicular side opposite to âˆ Î¸ / Base side adjacent to âˆ Î¸ â€¦â€¦ (2)

On comparing equation (1) and (2), we have

Perpendicular side opposite to âˆ Î¸Â = 12

Base side adjacent toÂ âˆ Î¸ = 13

Thus, in the triangle representingÂ âˆ Î¸Â we have,

Hypotenuse AC is the unknown and it can be found by using Pythagoras theorem

So by applying Pythagoras theorem, we have

AC^{2Â }= 12^{2}Â + 13^{2}

ACÂ ^{2}Â = 144 + 169

AC^{2}Â = 313

â‡’ AC = âˆš313

By definition,

sin Î¸ = Perpendicular side opposite to âˆ Î¸ / Hypotenuse = AB / AC

â‡’ sin Î¸ = 12/ âˆš313â€¦..(3)

And, cos Î¸ = Base side adjacent to âˆ Î¸ / Hypotenuse = BC / AC

â‡’ cos Î¸ = 13/ âˆš313 â€¦..(4)

Now, substituting the value ofÂ sin Î¸Â andÂ cos Î¸Â from equation (3) and (4) respectively in the equation below

Therefore,

### RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.2 Page No: 5.41

**Evaluate each of the following: **

**1. sinÂ 45 ^{âˆ˜}Â sinÂ 30^{âˆ˜}Â + cosÂ 45^{âˆ˜}Â cosÂ 30^{âˆ˜}**

**Solution:**

**2. sinÂ 60 ^{âˆ˜}Â cosÂ 30^{âˆ˜}Â + cosÂ 60^{âˆ˜}Â sinÂ 30^{âˆ˜}**

**Solution: **

**3. Â cosÂ 60 ^{âˆ˜}Â cosÂ 45^{âˆ˜}Â â€“ sinÂ 60^{âˆ˜}Â sinÂ 45^{âˆ˜}**

**Solution: **

**4. sin ^{2 }30^{âˆ˜} + sin^{2 }45^{âˆ˜} + sin^{2 }60^{âˆ˜} + sin^{2 }90^{âˆ˜}**

**Solution: **

**5. Â cos ^{2 }30^{âˆ˜} + cos^{2} 45^{âˆ˜} + cos^{2 }60^{âˆ˜} + cos^{2 }90^{âˆ˜}**

**Solution: **

**6. tan ^{2 }30^{âˆ˜} + tan^{2} 45^{âˆ˜} + tan^{2} 60^{âˆ˜}**

**Solution: **

**7. 2sin ^{2} 30^{âˆ˜} âˆ’ 3cos^{2 }45^{âˆ˜} + tan^{2 }60^{âˆ˜}**

**Solution: **

**8. sin ^{2} 30^{âˆ˜} cos^{2}45^{âˆ˜} + 4tan^{2} 30^{âˆ˜} + (1/2) sin^{2} 90^{âˆ˜} âˆ’ 2cos^{2} 90^{âˆ˜} + (1/24) cos20^{âˆ˜}**

**Solution: **

**9. 4(sin ^{4} 60^{âˆ˜} + cos^{4 }30^{âˆ˜}) âˆ’ 3(tan^{2 }60^{âˆ˜ }âˆ’ tan^{2} 45^{âˆ˜}) + 5cos^{2} 45^{âˆ˜}**

**Solution: **

**10. (cosec ^{2} 45^{âˆ˜} sec^{2 }30^{âˆ˜})(sin^{2} 30^{âˆ˜} + 4cot^{2} 45^{âˆ˜} âˆ’ sec^{2} 60^{âˆ˜})**

**Solution: **

**11. Â cosec ^{3} 30^{âˆ˜} cos60^{âˆ˜} tan^{3} 45^{âˆ˜} sin^{2} 90^{âˆ˜} sec^{2} 45^{âˆ˜} cot30^{âˆ˜}**

**Solution: **

**12. cot ^{2} 30^{âˆ˜} âˆ’ 2cos^{2} 60^{âˆ˜} âˆ’ (3/4)sec^{2} 45^{âˆ˜} â€“ 4sec^{2} 30^{âˆ˜}**

**Solution: **

Using trigonometric values, we have

**13. (cos0 ^{âˆ˜} + sin45^{âˆ˜} + sin30^{âˆ˜})(sin90^{âˆ˜} âˆ’ cos45^{âˆ˜} + cos60^{âˆ˜})**

**Solution: **

(cos0^{âˆ˜} + sin45^{âˆ˜} + sin30^{âˆ˜})(sin90^{âˆ˜} âˆ’ cos45^{âˆ˜} + cos60^{âˆ˜})

Using trigonometric values, we have

**15. 4/cot ^{2} 30^{âˆ˜} + 1/sin^{2} 60^{âˆ˜} âˆ’ cos^{2} 45^{âˆ˜}**

**Solution: **

**16. 4(sin ^{4 }30^{âˆ˜} + cos^{2} 60^{âˆ˜}) âˆ’ 3(cos^{2} 45^{âˆ˜} âˆ’ sin^{2} 90^{âˆ˜}) âˆ’ sin^{2} 60^{âˆ˜}**

**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 = âˆš3**

**Solution: **

Given,

2 sin 3x = âˆš3

sin 3x = âˆš3/2

sin 3x = sin 60Â°

3x = 60Â°

x = 20Â°

### RD Sharma Solutions for Class 10 Maths Chapter 5 Exercise 5.3 Page No: 5.52

**1. Evalute the following: **

**(i) sin 20 ^{o}/ cos 70^{o}**

**(ii) cos 19 ^{o}/ sin 71^{o}**

**(iii) sin 21 ^{o}/ cos 69^{o}**

**(iv) tan 10 ^{o}/ cot 80^{o}**

**(v) sec 11 ^{o}/ cosec 79^{o}**

**Solution:**

(i) We have,

sin 20^{o}/ cos 70^{o} = sin (90^{o} â€“ 70^{o})/ cos 70^{o }= cos 70^{o}/ cos70^{o} = 1 [âˆµ sin (90 – Î¸) = cos Î¸]

(ii) We have,

cos 19^{o}/ sin 71^{o }= cos (90^{o} â€“ 71^{o})/ sin 71^{o }= sin 71^{o}/ sin 71^{o} = 1 [âˆµ cos (90 – Î¸) = sin Î¸]

(iii) We have,

sin 21^{o}/ cos 69^{o }= sin (90^{o} â€“ 69^{o})/ cos 69^{o} = cos 69^{o}/ cos69^{o} = 1 [âˆµ sin (90 – Î¸) = cos Î¸]

(iv) We have,

tan 10^{o}/ cot 80^{o }= tan (90^{o} â€“ 10^{o}) / cot 80^{o} = cot 80^{o}/ cos80^{o} = 1 [âˆµ tan (90 – Î¸) = cot Î¸]

(v) We have,

sec 11^{o}/ cosec 79^{o} = sec (90^{o} â€“ 79^{o})/ cosec 79^{o} = cosec 79^{o}/ cosec 79^{o} = 1

**2. Evaluate the following: **

**Solution: **

We have, [âˆµ sin (90 – Î¸) = cos Î¸ and cos (90 – Î¸) = sin Î¸]

= 1^{2} + 1^{2} = 1 + 1

= 2

**(ii) cos 48Â°- sin 42Â°**

**Solution: **

We know that, cos (90Â° âˆ’ Î¸) = sin Î¸.

So,

cos 48Â°Â –Â sin 42Â°Â =Â cos (90Â° âˆ’ 42Â°) â€“ sin 42Â°Â = sin 42Â°Â –Â sin 42Â°= 0

Thus the value ofÂ cos 48Â°Â –Â sin 42Â°Â is 0.

**Solution: **

We have, [âˆµ cot (90 – Î¸) = tan Î¸ and cos (90 – Î¸) = sin Î¸]

= 1 â€“ 1/2(1)

= 1/2

**Solution: **

We have, [âˆµ sin (90 – Î¸) = cos Î¸ and cos (90 – Î¸) = sin Î¸]

= 1 â€“ 1

= 0

**Solution: **

We have, [âˆµ cot (90 – Î¸) = tan Î¸ and tan (90 – Î¸) = cot Î¸]

= tan (90^{o} â€“ 35^{o})/ cot 55^{o} + cot (90^{o} â€“ 12^{o})/ tan 12^{o} â€“ 1

= cot 55^{o}/ cot 55^{o} + tan 12^{o}/ tan 12^{o} â€“ 1

= 1 + 1 â€“ 1

= 1

**Solution: **

We have , [âˆµ sin (90 – Î¸) = cos Î¸ and sec (90 – Î¸) = cosec Î¸]

** **

= sec (90^{o} â€“ 20^{o})/ cosec 20^{o} + sin (90^{o} â€“ 31^{o})/ cos 31^{o}

= cosec 20^{o}/ cosec 20^{o} + cos 12^{o}/ cos 12^{o}

= 1 + 1

= 2

**(vii)Â cosec 31Â° – sec 59Â°**

**Solution:**

We have,

cosec 31Â° – sec 59Â°

Since, cosec (90 – Î¸) = cos Î¸

So,

cosec 31Â° – sec 59Â° = cosec (90Â° – 59^{o}) â€“ sec 59Â° = sec 59Â° – sec 59Â° = 0

Thus,

cosec 31Â° – sec 59Â° = 0

**(viii) (sin 72Â° + cos 18Â°) (sin 72Â° – cos 18Â°)**

**Solution: **

We know that,

sin (90 – Î¸) = cos Î¸

So, the given can be expressed as

(sin 72Â° + cos 18Â°) (sin (90 – 18)Â° – cos 18Â°)

= (sin 72Â° + cos 18Â°) (cos 18Â° – cos 18Â°)

= (sin 72Â° + cos 18Â°) x 0

= 0

**(ix) sin 35Â° sin 55Â° – cos 35Â° cos 55Â°**

**Solution: **

We know that,

sin (90 – Î¸) = cos Î¸

So, the given can be expressed as

sin (90 – 55)Â° sin (90 – 35)Â° – cos 35Â° cos 55Â°

= cos 55Â° cos 35Â° – cos 35Â° cos 55Â°

= 0

**(x) tan 48Â° tan 23Â° tan 42Â° tan 67Â°**

**Solution: **

We know that,

tan (90 – Î¸) = cot Î¸

So, the given can be expressed as

tan (90 – 42)Â° tan (90 – 67)Â° tan 42Â° tan 67Â°

= cot 42Â° cot 67Â° tan 42Â° tan 67Â°

= (cot 42Â° tan 42Â°)(cot 67Â° tan 67Â°)

= 1 x 1 [âˆµ tan Î¸ x cot Î¸ = 1]

= 1

**(xi) sec 50Â° sin 40Â° + cos 40Â° cosec 50Â°**

**Solution: **

We know that,

sin (90 – Î¸) = cos Î¸ and cos (90 – Î¸) = sin Î¸

So, the given can be expressed as

sec 50Â° sin (90 – 50)Â° + cos (90 – 50)Â° cosec 50Â°

= sec 50Â° cos 50Â° + sin 50Â° cosec 50Â°

= 1 + 1 [âˆµ sin Î¸ x cosec Î¸ = 1 and cos Î¸ x sec Î¸ = 1]

= 2

**3. Express each one of the following in terms of trigonometric ratios of angles lying between 0 ^{o} and 45^{o}**

**(i) sin 59 ^{o} + cos 56^{o} (ii) tan 65^{o} + cot 49^{o} (iii) sec 76^{o} + cosec 52^{o}**

**(iv) cos 78 ^{o} + sec 78^{o} (v) cosec 54^{o} + sin 72^{o} (vi) cot 85^{o} + cos 75^{o}**

**(vii) sin 67 ^{o} + cos 75^{o}**

**Solution: **

Using the below trigonometric ratios of complementary angles, we find the required

sin (90 – Î¸) = cos Î¸ cosec (90 – Î¸) = sec Î¸

cos (90 – Î¸) = sin Î¸ sec (90 – Î¸) = cosec Î¸

tan (90 – Î¸) = cot Î¸ cot (90 – Î¸) = tan Î¸

(i) sin 59^{o} + cos 56^{o} = sin (90 – 31)^{o} + cos (90 – 34)^{o} = cos 31^{o} + sin 34^{o}

(ii) tan 65^{o} + cot 49^{o} = tan (90 – 25)^{o} + cot (90 -41)^{o} = cot 25^{o} + tan 41^{o}

(iii) sec 76^{o} + cosec 52^{o} = sec (90 – 14)^{o} + cosec (90 – 38)^{o} = cosec 14^{o} + sec 38^{o}

(iv) cos 78^{o} + sec 78^{o }= cos (90 – 12)^{o} + sec (90 – 12)^{o} = sin 12^{o} + cosec 12^{o}

(v) cosec 54^{o} + sin 72^{o} = cosec (90 – 36)^{o} + sin (90 – 18)^{o} = sec 36^{o} + cos 18^{o}

(vi) cot 85^{o} + cos 75^{o} = cot (90 – 5)^{o} + cos (90 – 15)^{o} = tan 5^{o} + sin 15^{o}

**4. Express cos 75 ^{o} + cot 75^{o} in terms of angles between 0^{o} and 30^{o}. **

**Solution: **

Given,

cos 75^{o} + cot 75^{o}

Since, cos (90 – Î¸) = sin Î¸ and cot (90 – Î¸) = tan Î¸

cos 75^{o} + cot 75^{o } = cos (90 – 15)^{o} + cot (90 – 15)^{o} = sin 15^{o }+ tan 15^{o }

Hence, cos 75^{o} + cot 75^{o} can be expressed as sin 15^{o }+ tan 15^{o }

** 5. If sin 3A = cos (A â€“ 26 ^{o}), where 3A is an acute angle, find the value of A. **

**Solution: **

Given,

sin 3A = cos (A â€“ 26^{o})

Using cos (90 – Î¸) = sin Î¸, we have

sin 3A = sin (90^{o} – (A â€“ 26^{o}))

Now, comparing both L.H.S and R.H.S

3A = 90^{o} – (A â€“ 26^{o})

3A + (A â€“ 26^{o}) = 90^{o}

4A â€“ 26^{o} = 90^{o}

4A = 116^{o}

A = 116^{o}/4

âˆ´ A = 29^{o}

**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 = 180^{o}

So, C + A = 180^{o} â€“ B â‡’ (C + A)/2 = 90^{o} â€“ B/2 â€¦â€¦ (i)

And, B + C = 180^{o }â€“ A â‡’ (B + C)/2 = 90^{o} â€“ A/2 â€¦â€¦. (ii)

(i) L.H.S = tan ((C + A)/ 2)

â‡’ tan ((C + A)/ 2) = tan (90^{o} â€“ B/2) [From (i)]

= cot (B/2) [âˆµ tan (90 – Î¸) = cot Î¸]

= R.H.S

- Hence Proved

(ii) L.H.S = sin ((B + C)/2)

â‡’ sin ((B + C)/ 2) = sin (90^{o} â€“ A/2) [From (ii)]

= cos (A/2)

= R.H.S

- Hence Proved

**7. Prove that: **

**(i) tan 20Â° tan 35Â° tan 45Â° tan 55Â° tan 70Â° = 1**

**(ii) sin 48Â° sec 48Â° + cos 48Â° cosec 42Â° = 2**

**Solution: **

(i) Taking L.H.S = tan 20Â° tan 35Â° tan 45Â° tan 55Â° tan 70Â°

= tan (90Â° âˆ’ 70Â°) tan (90Â° âˆ’ 55Â°) tan 45Â°tan 55Â° tan70Â°

= cot 70Â°cot 55Â° tan 45Â° tan 55Â° tan 70Â°Â [âˆµ tan (90 – Î¸) = cot Î¸]

= (tan 70Â°cot 70Â°)(tan 55Â°cot 55Â°) tan 45Â°Â [âˆµ tan Î¸ x cot Î¸ = 1]

= 1 Ã— 1 Ã— 1 = 1

- Hence proved

(ii) Taking L.H.S = sin 48Â° sec 48Â° + cos 48Â° cosec 42Â°

= sin 48Â° sec (90Â° âˆ’ 48Â°) + cos 48Â° cosec (90Â° âˆ’ 48Â°)

[âˆµsec (90 – Î¸) = cosec Î¸ and cosec (90 – Î¸) = sec Î¸]= sin 48Â°cosec 48Â° + cos 48Â°sec 48Â°Â [âˆµ cosec Î¸ x sin Î¸ = 1 and cos Î¸ x sec Î¸ = 1]

= 1 + 1 = 2

- Hence proved

(iii) Taking the L.H.S,

= 1 + 1 â€“ 2

= 2 â€“ 2

= 0

- Hence proved

(iv) Taking L.H.S,

= 1 + 1

= 2

- Hence proved

**8. Prove the following: **

**(i) sinÎ¸ sin (90 ^{o} – Î¸) â€“ cos Î¸ cos (90^{o} – Î¸) = 0 **

**Solution: **

Taking the L.H.S,

sinÎ¸ sin (90^{o} – Î¸) â€“ cos Î¸ cos (90^{o} – Î¸)

= sin Î¸ cos Î¸ – cos Î¸ sin Î¸ [âˆµ sin (90 – Î¸) = cos Î¸ and cos (90 – Î¸) = sin Î¸]

= 0

**(ii)**

**Solution: **

Taking the L.H.S,

[âˆµ cosec Î¸ x sin Î¸ = 1 and cos Î¸ x sec Î¸ = 1]

= 1 + 1

= 2 = R.H.S

- Hence Proved

**(iii)**

**Solution: **

Taking the L.H.S, [âˆµ tan (90^{o} – Î¸) = cot Î¸]

= 0 = R.H.S

- Hence Proved

**(iv)**

**Solution: **

Taking L.H.S, [âˆµ sin (90 – Î¸) = cos Î¸ and cos (90 – Î¸) = sin Î¸]

= sin^{2} A = R.H.S

- Hence Proved

**(v) sin (50 ^{o} + **Î¸

**) – cos (40**Î¸

^{o}–**) + tan 1**

^{o}tan 10^{o}Â tan 20^{o}tan 70^{o}tan 80^{o}tan 89^{o}= 1**Solution: **

Taking the L.H.S,

= sin (50^{o} + Î¸) – cos (40^{o} – Î¸) + tan 1^{o} tan 10^{o} tan 20^{o} tan 70^{o} tan 80^{o} tan 89^{o}

= [sin (90^{o} â€“ (40^{o} – Î¸))]Â – cos (40^{o} – Î¸) + tan (90 – 89)^{o} tan (90 – 80)^{o} tan (90 – 70)^{o} tan 70^{o} tan 80^{o} tan 89^{o } [âˆµ sin (90 – Î¸) = cos Î¸]

= cos (40^{o}Â – Î¸) – cos (40^{o}Â – Î¸) + cot 89^{o} cot 80^{o} cot 70^{o} tan 70^{o} tan 80^{o} tan 89^{o}

^{o}– Î¸) = cot Î¸]

= 0 + (cot 89^{o} x tan 89^{o}) (cot 80^{o} x tan 80^{o}) (cot 70^{o} x tan 70^{o})

= 0 + 1 x 1 x 1 [âˆµ tan Î¸ x cot Î¸ = 1]

= 1= R.H.S

- Hence Proved