The science of measuring triangles is trigonometry. In other words, it deals with the measurement of sides and angles of triangles. This exercise contains problems dealing with finding all the trigonometric ratios when one of it is given. Further, the value of an expression is also asked to be evaluated. All the solutions to this and other chapters are easily available on the RD Sharma Solutions Class 10. For specific doubts regarding the solution and concept, students can download RD Sharma Solutions for Class 10 Maths Chapter 5 Trigonometric Ratios Exercise 5.1 PDF below.

## RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.1 Download PDF

### Access RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Exercise 5.1

**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/HypotenuseÂ

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¸Â

Also,Â secÂ Î¸ = Hypotenuse/Base

âˆ´ sec Î¸ = 17/15

Since, cot Î¸ = 1/tan Î¸Â

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¸Â

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¸Â

Also,Â secÂ Î¸ = Hypotenuse/Base

âˆ´ sec Î¸ = 13/12

Since,Â tanÎ¸ = 1/cot Î¸Â

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¸Â

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¸Â

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/PQÂ

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 RÂ

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¸ = 1/3……. (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

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

Â

**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,Â

Â

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

**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,