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

Question 7. If cotθ = 7/8, evaluate:

(i) 

(ii) cot²θ 

Solution:

 cotθ = 7/8 = Base/Perpendicular

In right-angled ΔPQR,

∠Q = 90°, PQ = 8, RQ = 7

Using Pythagoras Theorem

PR² = PQ² + QR²

PR² = 8² + 7² = 64 + 49

PR² = 113

PR = √113
Now

sinθ = 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) cot²θ  

= (cosθ/sinθ)²

Putting the values of sinθ and cosθ in the following equation

=

=

= 49/64

or

cot²θ = (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

AC² = AB² + BC²

AC² = 4² + 3² = 16 + 9

AC² = 25

AC = 5

Now, 

Taking LHS

=

=

=

=

= 7/25

Taking RHS

= cos²A – sin²A

=(

=(

=(

= 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

AC² = BC² + AB²

AC² = a² + b²

AC² = 

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

AC² = BC² + AB²

AC² = 3² + 4²

AC² = 9 + 16 = 25

AC = 5

Now, 

=

= 6/18 = 1/3 

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

Solution:  

Given, tanθ = a/b 

Using Pythagoras Theorem

AC² = BC² + AB²

AC² = a² + b²

AC² = 

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

AC² = BC² + AB²

13² = BC² + 5²

BC² = 169 – 25 = 144

BC = 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

AC² = BC² + AB²

13² = BC² + 12²

BC² = 169 – 144 = 25

BC = 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,

AC² = AB² + BC²

AC² = 1² + (√3)²

AC² = 3 + 1 = 4

AC = 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 sec²θ = (1 + tan²θ) = 1 + 1/7 = 8/7

and cosec²θ = (1 + cot²θ) = 1 + 7 = 8

Now, 

=

= 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,

AC² = BC² + AB²

5² = BC² + 4²

BC² = 25 − 16 = 9

BC = 3

Now,

=

=

= 12/7   

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

Solution:

Given: tanθ = 12/13

From Pythagoras theorem,

AC² = BC² + AB²

AC² = (13)² + (12)²

AC² = 313

AC = √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,

AC² = BC² + AB²

5² = 3² + AB²

AB² = 25 − 9 = 16

AB = 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,

AC² = BC² + AB²

AC² = 24² + 7²

AC² = 576 + 49 = 625

AC = 25

Now, 

= sinθ + cosθ

= 24/25 + 7/25

= (24 + 7)/25

= 31/25