# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.1 | Set 2

### Question 7. If cotθ = 7/8, evaluate:

### (i)

### (ii) cot^{2}θ

**Solution:**

cotθ = 7/8 = Base/Perpendicular

In right-angled ΔPQR,

∠Q = 90°, PQ = 8, RQ = 7

Using Pythagoras Theorem

PR

^{2 }= PQ^{2 }+ QR^{2}PR

^{2 }= 8^{2 }+ 7^{2 }= 64 + 49PR

^{2 }= 113PR = √113

Nowsinθ = Perpendicular/Hypotenuse = PQ/PR = 8/√113

cosθ = Base/Hypotenuse = QR/PR = 7/√113

(i)Putting the values of sinθ and cosθ in the equation

=

=

= 49/64

(ii)cot^{2}θ= (cosθ/sinθ)

^{2}Putting the values of sinθ and cosθ in the following equation

=

=

= 49/64

or

cot

^{2}θ = (cotθ)^{2 }= (7/8)^{2 }= 49/64

### Question 8. If 3 cot A = 4, check whether or not.

**Solution:**

Given, 3cot A = 4 or cot A = 4/3

Draw a △ ABC where ∠B = 90°, AB = 4, BC = 3

Using Pythagoras Theorem

AC

^{2 }= AB^{2 }+ BC^{2}AC

^{2 }= 4^{2 }+ 3^{2 }= 16 + 9AC

^{2 }= 25AC = 5

Now,

Taking LHS

=

=

=

=

= 7/25

Taking RHS

= cos

^{2}A – sin^{2}A=(

=(

=(

= 7/25

RHS = LHS (Hence Proved)

### Question 9. If tanθ = a/b, find the value of

**Solution:**

Given, tanθ = a/b

Draw a △ABC where ∠B = 90°, AB = b, BC = a

Using Pythagoras Theorem

AC

^{2 }= BC^{2 }+ AB^{2}AC

^{2 }= a^{2 }+ b^{2}AC

^{2 }=Now,

=

=

= (b + a)/(b – a)

### Question 10. If 3tanθ = 4, find the value of

**Solution:**

Given tanθ = 4/3

Now, Dividing the numerator and denominator by cosθ

=

Putting the values of tanθ in the above equation

=

=

= 8/10

= 4/5

### Question 11. If 3cotθ = 2, find the value of

**Solution:**

Given: 3cotθ = 2

Using Pythagoras Theorem

AC

^{2 }= BC^{2}+ AB^{2}AC

^{2 }= 3^{2 }+ 4^{2}AC

^{2 }= 9 + 16 = 25AC = 5

Now,

=

= 6/18 = 1/3

### Question 12. If tanθ = a/b, prove that

**Solution: **

Given, tanθ = a/b

Using Pythagoras Theorem

AC

^{2 }= BC^{2 }+ AB^{2}AC

^{2 }= a^{2 }+ b^{2}AC

^{2 }=Now,

=

Putting the values of sinθ and cosθ in the above equation

=

=

=

=

Hence Proved

### Question 13. If secθ = 13/5, prove that =3

**Solution: **

Given, secθ = 13/5

Using Pythagoras Theorem

AC

^{2 }= BC^{2}+ AB^{2}13

^{2 }= BC^{2 }+ 5^{2}BC

^{2 }= 169 – 25 = 144BC = 12

Now,

Taking LHS

=

Putting the values of sinθ and cosθ in the above equation

= 3 = RHS

Hence Proved

### Question 14. If cosθ = 12/13, show that sinθ(1 – tanθ) = 35/156

**Solution:**

We have cosθ = 12/13

Using Pythagoras Theorem

AC

^{2 }= BC^{2 }+ AB^{2}13

^{2 }= BC^{2 }+ 12^{2}BC

^{2 }= 169 – 144 = 25BC = 5

Now,

Taking LHS

= sinθ(1 – tanθ)

=

=

= 35/156

= RHS

Hence Proved

### Question 15. If cotθ = 1/√3, show that

**Solution:**

Given, cotθ = 1/√3

tanθ = 1/cotθ =√3

From Pythagoras theorem,

AC

^{2 }= AB^{2 }+ BC^{2}AC

^{2 }= 1^{2 }+ (√3)^{2}AC

^{2 }= 3 + 1 = 4AC = 2

Now,

Taking LHS

=

=

= 3/5

Hence Proved

### Question 16. If tanθ = 1/√7, then

**Solution:**

We have

tanθ = 1/√7

cotθ = √7

We know sec

^{2}θ = (1 + tan^{2}θ) = 1 + 1/7 = 8/7and cosec

^{2}θ = (1 + cot^{2}θ) = 1 + 7 = 8Now,

=

= 48/64 = 3/4

### Question 17. If secθ = 5/4, find the value of

**Solution:**

Given:

secθ = 5/4

cosθ = 1/secθ = 4/5

From Pythagoras theorem,

AC

^{2 }= BC^{2 }+ AB^{2}5

^{2 }= BC^{2 }+ 4^{2}BC

^{2 }= 25 − 16 = 9BC = 3

Now,

=

=

= 12/7

### Question 18. If tanθ = 12/13, find the value of

**Solution:**

Given: tanθ = 12/13

From Pythagoras theorem,

AC

^{2 }= BC^{2 }+ AB^{2}AC

^{2 }= (13)^{2}+ (12)^{2}AC

^{2 }= 313AC = √313

sinθ = 12/√313

cosθ = 13/√313

We have

= \

=

= 312/25

### Question 19. If cosθ = 3/5, then evaluate

**Solution:**

Given:

cosθ = 3/5

From Pythagoras theorem,

AC

^{2 }= BC^{2 }+ AB^{2}5

^{2 }= 3^{2 }+ AB^{2}AB

^{2 }= 25 − 9 = 16AB = 4

Now

=

=

=

= (1/20) × (3/8)

= 3/160

### Question 20**. If **sinθ = 3/5,** then evaluate**

**Solution:**

Given,

sinθ = 3/5

Now

=

=

=

=

=

= (sinθ – 1)/(2)

Putting the value of sinθ, we get

=

=

= -1/5

### Question 21. If tanθ = 24/7, find that sinθ + cosθ.

**Solution:**

Given:

tanθ = 24/7

From Pythagoras theorem,

AC

^{2 }= BC^{2 }+ AB^{2}AC

^{2 }= 24^{2 }+ 7^{2}AC

^{2 }= 576 + 49 = 625AC = 25

Now,

= sinθ + cosθ

= 24/25 + 7/25

= (24 + 7)/25

= 31/25