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

Trigonometry is a fundamental branch of mathematics that deals with the relationships between the angles and sides of triangles. Chapter 5 of RD Sharma's Class 10 textbook focuses on the Trigonometric Ratios which are the foundation of trigonometry. In this chapter, students will learn to apply these ratios to solve various problems involving the right-angled triangles. This exercise provides the practice problems that will help solidify the understanding of the trigonometric concepts.

Trigonometric Ratios

The Trigonometric Ratios are the ratios of the sides of a right-angled triangle with the respect to its angles. The primary ratios are sine, cosine, and tangent which are defined as the ratio of the opposite side to the hypotenuse adjacent side to the hypotenuse and opposite side to the adjacent side respectively. These ratios are essential in, solving problems related to angles and distances.

Question 22. If sinθ = a/b, find secθ + tanθ in terms of a and b.

Solution:

Given:

sinθ = a/b

From Pythagoras theorem,

AC² = BC² + AB²

b² = a² + AB²

AB²=\sqrt{(b^2-a^2)}     

Now, 

= secθ + tanθ

=\frac{b}{\sqrt(b^2-a^2)}+\frac{a}{\sqrt(b^2-a^2)}

=\frac{b+a}{\sqrt(b^2-a^2)}

=\frac{(\sqrt(b+a))^2}{\sqrt(b-a)\sqrt(b+a)}

=\frac{\sqrt(b+a)}{\sqrt(b-a)}

=\sqrt\frac{b+a}{b-a}

Question 23. If 8tanA = 15, find sin A − 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 − cos A

= 15/17 - 8/17

= (15 - 8)/17

= 7/17  

Question 24. If tanθ = 20/21, show that \frac{1−sinθ+cosθ}{1+sinθ+cosθ}=\frac{3}{7}  .

Solution:

Given: tanθ = 20/21

From Pythagoras theorem,

AC² = BC² + AB²

AC² = 20² + 21²

AC² = 400 + 441 = 841

AC = 29

Now, taking LHS

=\frac{1−sinθ+cosθ}{1+sinθ+cosθ}

=\frac{1−\frac{20}{29}+\frac{21}{29}}{1+\frac{20}{29}+\frac{21}{29}}

=\frac{29-20+21}{29+20+21}

= 30/70

= 3/7

Question 25. If cosec A = 2, find the value of \frac{1}{tanA}+\frac{sinA}{1+cosA}  .

Solution:

Given:

cosec A = 2

We know 

sin A = 1/cosecA = 1/2

And, sin 30° = 1/2

A = 30°

tan30° = 1/√3 and cos30° = √3/2 

Now,

= \frac{1}{tanA}+\frac{sinA}{1+cosA}

= \frac{1}{\frac{1}{\sqrt3}}+\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}

= \sqrt3+\frac{\frac{1}{2}}{\frac{2+\sqrt3}{2}}

= \sqrt3+\frac{1}{2+\sqrt3}

= \frac{2\sqrt3+3+1}{2+\sqrt3}   
= \frac{2\sqrt3+4}{2+\sqrt3}     

= 2 

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

Solution:

Let us consider a △ABC 

From the figure,

Given,

cos A = cos B

AC/AB = BC/AB

Multiplying both side by AB

(AC/AB) × AB = (BC/AB) × AB 

AC = BC

In △ABC, 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, ∠A =∠B

Question 27. In a Δ ABC, right angled at A, if tanC = √3, find the value of sin B cos C + cos B sin C.

Solution:

In right angled Δ ABC,

Given: tan C = √3 

∴AB = √3 and AC = 1

From pythagoras theorem,

BC² = AB² + AC²

BC² = (√3)² + 1²

BC² = 3 + 1 = 4

BC = 2

Therefore,

sin B cos C + cos B sin C

= (1/2)(√3/2) + (√3/2)(√3/2)

= 1/4 + 3/4

= 4/4

= 1

Question 28. State whether the following are true or false. Justify your answer.

(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 ∠A

secθ = Hypotenuse/Base 

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

So secθ > 1 is valid 

Here, secθ = 12/5  

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

Solution:

FALSE 

cos A means cosine of ∠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 ∠A

cot A = 1/tanA

Only "cot" doesn't defines anything.

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

(v)  sinθ = 4/3 for some angle θ.

Solution:

FALSE

sinθ = 4/3 for some value of ∠θ

We have,

sinθ = Perpendicular/Hypotenuse 

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

So sinθ is always less than 1.

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

Question 29. If sinθ = 12/13, find the value of \frac{sin^2θ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}  .

Solution:

Given:

sinθ = 12/13

Using Pythagoras theorem,

AC² = BC² + AB²

13² = 12² + AB²

AB² = 169 − 144 = 25

AB = 5

=  \frac{sin^2θ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}

= \frac{(\frac{12}{13})^2-(\frac{5}{13})^2}{2(\frac{12}{13})(\frac{5}{13})}×\frac{1}{(\frac{12}{5})^2}

= \frac{\frac{144}{169}-\frac{25}{169}}{\frac{120}{169}}×\frac{25}{144}

= \frac{\frac{144-25}{169}}{\frac{120}{169}}×\frac{25}{144}

= \frac{119}{120}×\frac{25}{144}

= 595/3456 

Question 30. If cosθ = 5/13, find the value of \frac{sinθ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}  .

Solution:

Given:

cosθ = 5/13

Using Pythagoras theorem,

AC² = BC² + AB²

13² = BC² + 52

BC² = 169 − 25 = 144

BC = 12

= \frac{sinθ-cos^2θ}{2sinθcosθ}×\frac{1}{tan^2θ}

= \frac{(\frac{12}{13})-(\frac{5}{13})^2}{2(\frac{12}{13})(\frac{5}{13})}×\frac{1}{(\frac{12}{5})^2}

= \frac{\frac{144}{169}-\frac{25}{169}}{\frac{120}{169}}×\frac{25}{144}

= \frac{\frac{144-25}{169}}{\frac{120}{169}}×\frac{25}{144}

= \frac{119}{120}×\frac{25}{144}

= 595/3456 

Question 31. If secA = 5/4, verify that \frac{3sinA−4sin^3A }{4cos^3A−3cosA}=\frac{3tanA−tan^3A}{1−3tan^2A}  .

Solution:

Given:

 secA = 5/4

From pythagoras theorem,

AC² = BC² + AB²

5² = BC² + 4²

BC² = 25 − 16 = 9

BC = 3

Now 

=\frac{3sinA−4sin^3A }{4cos^3A−3cosA}=\frac{3tanA−tan^3A}{1−3tan^2A}

=\frac{3(\frac{3}{5})−4(\frac{3}{5})^3 }{4(\frac{4}{5})^3−3(\frac{4}{5})}=\frac{3(\frac{3}{4})−(\frac{3}{4})^3}{1−3(\frac{3}{4})^2}

=\frac{(\frac{9}{5})−(\frac{108}{125}) }{(\frac{256}{125})−(\frac{12}{5})}=\frac{(\frac{9}{4})−(\frac{27}{64})}{1−(\frac{27}{16})}

=\frac{(\frac{225-108}{125}) }{(\frac{256-300}{125})}=\frac{(\frac{144-27}{64})}{(\frac{16-27}{16})}

=\frac{(\frac{117}{125}) }{(\frac{-44}{125})}=\frac{(\frac{117}{64})}{(\frac{-11}{16})}

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

= 117/-44 = 117/-44

Hence Proved

Question 32. If sinθ = 3/4, prove that \sqrt\frac{cosec^2θ−cot^2θ}{sec^2θ−1}=\frac{\sqrt 7}{3}   .  

Solution:

Given: sinθ = 3/4 

From Pythagoras theorem,

AC² = BC² + AB²

4² = AB² + 3²

AB² = 16 - 9 = 7

AB =√7

We have,

\sqrt\frac{cosec^2θ−cot^2θ}{sec^2θ−1}=\frac{\sqrt 7}{3}

 Now squaring both side

=(\sqrt\frac{cosec^2θ−cot^2θ}{sec^2θ−1})^2=(\frac{\sqrt 7}{3})^2

=\frac{cosec^2θ−cot^2θ}{sec^2θ−1}   

= 7/9

We know

1 + cot²θ = cosec²θ

1 + tan²θ = sec²θ

= 1/tan²θ = 7/9 

=\frac{1}{(\frac{3}{\sqrt7})^2}=\frac{7}{9}

= 7/9 = 7/9 

Hence Proved

Question 33. If secA = 17/8, verify that \frac{3−4sin^2A}{4cos^2A−3}=\frac{3−tan^2A}{1−3tan^2A}  .

Solution:

Given: secA = 17/8

From Pythagoras theorem,

AC² = BC² + AB²

17² = BC² + 8²

BC² = 289 − 64 = 225

BC = 15\frac{(\frac{-33}{289})}{(\frac{-611}{289})}=\frac{(\frac{-33}{64})}{(\frac{-611}{64})}

We have 

\frac{3−4sin^2A}{4cos^2A−3}=\frac{3−tan^2A}{1−3tan^2A}

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

=\frac{3−4\frac{15}{17}^2}{4\frac{8}{17}^2−3}=\frac{3−\frac{15}{8}^2}{1−3\frac{15}{8}^2}

=\frac{3−(\frac{900}{289})}{(\frac{256}{289})−3}=\frac{3−(\frac{225}{64})}{1−(\frac{675}{64})}

=\frac{(\frac{867-900}{289})}{(\frac{256-867}{289})}=\frac{(\frac{192-225}{64})}{(\frac{64-675}{64})}

=\frac{(\frac{-33}{289})}{(\frac{-611}{289})}=\frac{(\frac{-33}{64})}{(\frac{-611}{64})}

= 33/611 = 33/611

Hence Proved

Question 34. If cotθ = 3/4, prove that \sqrt\frac{secθ−cosecθ}{secθ+cosecθ}=\frac{1}{\sqrt 7}   .    

Solution:

Given:  cotθ = 3/4

 tanθ = 4/3

Using the pythagoras theorem 

 sinθ = 4/5, cosθ = 3/5

cosecθ = 5/4, secθ = 5/3

Now, taking LHS

=\sqrt\frac{secθ−cosecθ}{secθ+cosecθ}  

=\sqrt\frac{\frac{5}{3}−\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}  

=\sqrt\frac{\frac{5}{12}}{\frac{35}{12}}  

=\sqrt\frac{5}{35}    

= 1/√7

Question 35. If 3cosθ − 4sinθ = 2cosθ + sinθ, then find tanθ.

Solution:

Given: 3cosθ − 4sinθ = 2cosθ + sinθ

Dividing both equation by cosθ we get,

3\frac{cosθ}{cosθ}−4\frac{sinθ}{cosθ}=2\frac{cosθ}{cosθ}+\frac{sinθ}{cosθ}

3 - 4tanθ = 2 + tanθ

3 - 2 = 4tanθ + tanθ

tanθ = 1/5

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

Solution:

Let us consider a △ABC  

From the figure,

Given:

tan A = tan B

BC/AC = AC/BC

AC² = BC²

AC = BC

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

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 ∠A =∠B

Read more :

• Trigonometric Identities
• Trigonometry in Maths: Table, Formulas, Identities and Ratios