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Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.1 | Set 3
  • Last Updated : 21 Feb, 2021

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 eqution 

=\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

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