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# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.1 | Set 3

• Last Updated : 04 Aug, 2021

### Question 22. If sinÎ¸ = a/b, find secÎ¸ + tanÎ¸ in terms of a and b.

Solution:

Given:

sinÎ¸ = a/b

From Pythagoras theorem,

AC2 = BC2 + AB2

b2 = a2 + AB2

AB2=

Now,

= secÎ¸ + tanÎ¸

=

=

=

=

=

### Question 23. If 8tanA = 15, find sin A âˆ’ cos A.

Solution:

Given:

8tanA = 15

tanA = 15/8

From pythagoras theorem,

AC2 = BC2 + AB2

AC2 = 152 + 82

AC2 = 225 + 64 = 289

AC = 17

Now,

= sin A âˆ’ cos A

= 15/17 – 8/17

= (15 – 8)/17

= 7/17

### Question 24. If tanÎ¸ = 20/21, show that .

Solution:

Given: tanÎ¸ = 20/21

From Pythagoras theorem,

AC2 = BC2 + AB2

AC2 = 202 + 212

AC2 = 400 + 441 = 841

AC = 29

Now, taking LHS

=

=

=

= 30/70

= 3/7

### Question 25. If cosec A = 2, find the value of .

Solution:

Given:

cosec A = 2

We know

sin A = 1/cosecA = 1/2

And, sin 30Â° = 1/2

A = 30Â°

tan30Â° = 1/âˆš3 and cos30Â° = âˆš3/2

Now,

=

=

=

=

=
=

= 2

### Question 26. If âˆ A and âˆ B are acute angles such that cos A = cos B, then show that âˆ A = âˆ B.

Solution:

Let us consider a â–³ABC

From the figure,

Given,

cos A = cos B

AC/AB = BC/AB

Multiplying both side by AB

(AC/AB) Ã— AB = (BC/AB) Ã— AB

AC = BC

In â–³ABC, AC = BC So we can say that the triangle is an isosceles triangle,

and in an isosceles triangle we know that if two sides of a triangle are equal,

then the angle opposite to the sides are equal.

Therefore, âˆ A =âˆ B

### Question 27. In a Î” ABC, right angled at A, if tanC = âˆš3, find the value of sin B cos C + cos B sin C.

Solution:

In right angled Î” ABC,

Given: tan C = âˆš3

âˆ´AB = âˆš3 and AC = 1

From pythagoras theorem,

BC2 = AB2 + AC2

BC2 = (âˆš3)2 + 12

BC2 = 3 + 1 = 4

BC = 2

Therefore,

sin B cos C + cos B sin C

= (1/2)(âˆš3/2) + (âˆš3/2)(âˆš3/2)

= 1/4 + 3/4

= 4/4

= 1

### (i) The value of tan A is always less than 1.

Solution:

FALSE. The value of tan A is not always less than 1.

Consider the Pythagorean triplet, 13, 12, and 5

where, 13 is the hypotenuse

We know

tan A = Perpendicular/Base

Let Perpendicular = 12 and Base = 5

then, tanA = 12/5 = 2.4 which is greater than 1.

### (ii) sec A = 12/5 for some value of angle A.

Solution:

TRUE

We have sec A = 12/5 for some value of âˆ A

secÎ¸ = Hypotenuse/Base

In a right angled triangle, hypotenuse is the greatest side.

So secÎ¸ > 1 is valid

Here, secÎ¸ = 12/5

### (iii) cos A is the abbreviation used for the cosecant of angle A.

Solution:

FALSE

cos A means cosine of âˆ A

cos A = Base/Hypotenuse

However,

cosec A = Hypotenuse/Perpendicular

### (iv) cot A is the product of cot and A.

Solution:

FALSE

cot A means Cotangent of âˆ A

cot A = 1/tanA

Only “cot” doesn’t defines anything.

Hence, cot A is not the product of cot and A.

### (v)  sinÎ¸ = 4/3 for some angle Î¸.

Solution:

FALSE

sinÎ¸ = 4/3 for some value of âˆ Î¸

We have,

sinÎ¸ = Perpendicular/Hypotenuse

In a right angled triangle, hypotenuse is the greatest side.

So sinÎ¸ is always less than 1.

Here, sinÎ¸ = 4/3 = 1.3 which is greater than 1

### Question 29. If sinÎ¸ = 12/13, find the value of.

Solution:

Given:

sinÎ¸ = 12/13

Using Pythagoras theorem,

AC2 = BC2 + AB2

132 = 122 + AB2

AB2 = 169 âˆ’ 144 = 25

AB = 5

=

=

=

=

= 595/3456

### Question 30. If cosÎ¸ = 5/13, find the value of .

Solution:

Given:

cosÎ¸ = 5/13

Using Pythagoras theorem,

AC2 = BC2 + AB2

132 = BC2 + 52

BC2 = 169 âˆ’ 25 = 144

BC = 12

=

=

= 595/3456

### Question 31. If secA = 5/4, verify that .

Solution:

Given:

secA = 5/4

From pythagoras theorem,

AC2 = BC2 + AB2

52 = BC2 + 42

BC2 = 25 âˆ’ 16 = 9

BC = 3

Now

=

=

=

=

=

= 117/-44 = 117/(11(4))

= 117/-44 = 117/-44

Hence Proved

### Question 32. If sinÎ¸ = 3/4, prove that .

Solution:

Given: sinÎ¸ = 3/4

From Pythagoras theorem,

AC2 = BC2 + AB2

42 = AB2 + 32

AB2 = 16 – 9 = 7

AB =âˆš7

We have,

Now squaring both side

=

=

= 7/9

We know

1 + cot2Î¸ = cosec2Î¸

1 + tan2Î¸ = sec2Î¸

= 1/tan2Î¸ = 7/9

=

= 7/9 = 7/9

Hence Proved

### Question 33. If secA = 17/8, verify that .

Solution:

Given: secA = 17/8

From Pythagoras theorem,

AC2 = BC2 + AB2

172 = BC2 + 82

BC2 = 289 âˆ’ 64 = 225

BC = 15

We have

Putting the values of sinA, cosA and tanA in the above equation

=

=

=

=

= 33/611 = 33/611

Hence Proved

### Question 34. If cotÎ¸ = 3/4, prove that .

Solution:

Given:  cotÎ¸ = 3/4

tanÎ¸ = 4/3

Using the pythagoras theorem

sinÎ¸ = 4/5, cosÎ¸ = 3/5

cosecÎ¸ = 5/4, secÎ¸ = 5/3

Now, taking LHS

=

=

=

=

= 1/âˆš7

### Question 35. If 3cosÎ¸ âˆ’ 4sinÎ¸ = 2cosÎ¸ + sinÎ¸, then find tanÎ¸.

Solution:

Given: 3cosÎ¸ âˆ’ 4sinÎ¸ = 2cosÎ¸ + sinÎ¸

Dividing both equation by cosÎ¸ we get,

3 – 4tanÎ¸ = 2 + tanÎ¸

3 – 2 = 4tanÎ¸ + tanÎ¸

tanÎ¸ = 1/5

### Question 36. If âˆ A and âˆ B are acute angles such that tan A = tan B, then show that âˆ A = âˆ B.

Solution:

Let us consider a â–³ABC

From the figure,

Given:

tan A = tan B

BC/AC = AC/BC

AC2 = BC2

AC = BC

In â–³ABC, AC = BC So we can say that the triangle is an isosceles triangle, âˆš3

and in an isosceles triangle we know that if two sides of a triangle are equal,

then the angle opposite to the sides are equal.

Therefore âˆ A =âˆ B

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