RD Sharma Solutions Class 10 Trigonometric Ratios Exercise 5.3

RD Sharma Solutions Class 10 Chapter 5 Exercise 5.3

RD Sharma Class 10 Solutions Chapter 5 Ex 5.3 PDF Free Download

Exercise 5.3

 

Q.1) Evaluate the following : (i) \(\frac{sin 20}{cos 70}\)

 

Sol (i) : Given that, \(\frac{sin 20}{cos 70}\)

Since sin (90 – \(\Theta\)) = cos\(\Theta\)

\(\Rightarrow\) \(\frac{sin 20}{cos 70}\)= \(\frac{sin (90-70)}{cos 70}\)

\(\Rightarrow\) \(\frac{sin 20}{cos 70}\) = \(\frac{cos 70}{cos 70}\)

\(\Rightarrow\) \(\frac{sin 20}{cos 70}\) =1

Therefore \(\frac{sin 20}{cos 70}\) = 1

 

(ii) \(\frac{cos 19}{sin 71}\)

 

Soln.(ii): Given that, \(\frac{cos 19}{sin 71}\)

\(\Rightarrow\) \(\frac{cos 19}{sin 71}\) = \(\frac{cos (90-71)}{sin 71}\)

\(\Rightarrow\)\(\frac{cos 19}{sin 71}\) = \(\frac{sin 71}{sin 71}\)

\(\Rightarrow\)\(\frac{cos 19}{sin 71}\) =1

Since cos(90-\(\Theta\))= sin \(\Theta\)

Therefore \(\frac{cos 19}{sin 71}\) = 1

 

(iii) \(\frac{sin21}{cos69}\)

 

Soln.(iii): Given that, \(\frac{sin21}{cos69}\)

Since (90-\(\Theta\)) = cos\(\Theta\)

\(\Rightarrow\)\(\frac{sin21}{cos69}\) = \(\frac{sin(90-69)}{cos69}\)

\(\Rightarrow\)\(\frac{sin21}{cos69}\) = \(\frac{cos69}{cos69}\)

\(\Rightarrow\)\(\frac{sin21}{cos69}\) =1

 

(iv) \(\frac{tan10}{cot80}\)

 

Soln.(iv): We are given that, \(\frac{tan10}{cot80}\)

Since tan(90-\(\Theta\)) = cot\(\Theta\)

\(\Rightarrow\)\(\frac{tan10}{cot80}\) =\(\frac{tan(90-80)}{cot80}\)

\(\Rightarrow\) \(\frac{tan10}{cot80}\) =

\(\frac{cot80}{cot80}\)

\(\Rightarrow\) \(\frac{tan10}{cot80}\) = 1

Therefore \(\frac{tan10}{cot80}\) = 1

 

(v) \(\frac{sec11}{cosec79}\)

 

Soln.(v):

Given that, \(\frac{sec11}{cosec79}\)

Since sec(90-\(\Theta\))=cosec\(\Theta\)

\(\Rightarrow\)\(\frac{sec11}{cosec79}\) = \(\frac{sec(90-79)}{cosec79}\)

\(\Rightarrow\)\(\frac{sec11}{cosec79}\) = \(\frac{cosec79}{cosec79}\)

\(\Rightarrow\)\(\frac{sec11}{cosec79}\) = 1

Therefore \(\frac{sec11}{cosec79}\) = 1

 

Q.2: EVALUATE THE FOLLOWING :

 

(i) \(\left ( \frac{sin49^{\circ}}{cos41^{\circ}} \right )^{2}\) + \(\left ( \frac{cos41^{\circ}}{sin49^{\circ}} \right )^{2}\)

 

Soln.(i):

We have to find: \(\left ( \frac{sin49^{\circ}}{cos41^{\circ}} \right )^{2}\) + \(\left ( \frac{cos41^{\circ}}{sin49^{\circ}} \right )^{2}\)

Since \(\frac{sec70^{\circ}}{cosec20^{\circ}}\) + \(\frac{sin59^{\circ}}{cos31^{\circ}}\) sin(\(90^{\circ}\)\(\Theta\)) = cos\(\Theta\) and cos(\(90^{\circ}\)\(\Theta\)) = sin\(\Theta\)

So

\(\left (\frac{sin(90^{\circ}-41^{\circ})}{cos41^{\circ}}\right )^{2}\) + \(\left (\frac{cos(90^{\circ}-49^{\circ})}{sin49^{\circ}}\right )^{2}\) = \(\left ( \frac{cos41^{\circ}}{cos41^{\circ}} \right )^{2}\) + \(\left ( \frac{sin49^{\circ}}{sin49^{\circ}} \right )^{2}\)

= 1+1 = 2

So value of \(\left ( \frac{sin49^{\circ}}{cos41^{\circ}} \right )^{2}\) + \(\left ( \frac{cos41^{\circ}}{sin49^{\circ}} \right )^{2}\) is 2

 

(ii) \(cos48^{\circ}\)\(sin42^{\circ}\)

 

Soln.(ii)

We have to find: \(cos48^{\circ}\)\(sin42^{\circ}\)

Since cos(\(90^{\circ}-\Theta\)) = sin\(\Theta\).So

\(cos48^{\circ}\)\(sin42^{\circ}\) = \(cos(90^{\circ}-42^{\circ})-sin42^{\circ}\)

=\(sin42^{\circ}\)\(sin42^{\circ}\) =0

So value of \(cos48^{\circ}\)\(sin42^{\circ}\) is 0

 

(iii) \(\frac{cot40^{\circ}}{tan50^{\circ}}-\frac{1}{2}\left ( \frac{cos35^{\circ}}{sin55^{\circ}} \right )\)

 

Soln.(iii)

We have to find:

\(\frac{cot40^{\circ}}{tan50^{\circ}}-\frac{1}{2}\left ( \frac{cos35^{\circ}}{sin55^{\circ}} \right )\)

Since cot(\(90^{\circ}-\Theta\)) = tan\(\Theta\) and cos(\(90^{\circ}-\Theta\))=sin\(\Theta\)

\(\frac{cot40^{\circ}}{tan50^{\circ}}-\frac{1}{2}\left ( \frac{cos35^{\circ}}{sin55^{\circ}} \right )\)= \(\frac{cot(90^{\circ}-50^{\circ})}{tan50^{\circ}}-\frac{1}{2}\left ( \frac{cos(90^{\circ}-55^{\circ})}{sin55^{\circ}} \right )\)

=\(\frac{tan50^{\circ}}{tan50^{\circ}}-\frac{1}{2}\left ( \frac{sin55^{\circ}}{sin55^{\circ}} \right )\)

= \(1-\frac{1}{2}\) = \(\frac{1}{2}\)

So value of \(\frac{cot40^{\circ}}{tan50^{\circ}}-\frac{1}{2}\left ( \frac{cos35^{\circ}}{sin55^{\circ}} \right )\) is \(\frac{1}{2}\)

 

(iv)\(\left ( \frac{sin27^{\circ}}{cos63^{\circ}} \right )^{2}\)\(\left ( \frac{cos63^{\circ}}{sin27^{\circ}} \right )^{2}\)

 

Soln(iv)

We have to find: \(\left ( \frac{sin27^{\circ}}{cos63^{\circ}} \right )^{2}\)\(\left ( \frac{cos63^{\circ}}{sin27^{\circ}} \right )^{2}\)

Since sin (\(90^{\circ}-\Theta\))= cos\(\Theta\) and cos (\(90^{\circ}-\Theta\))=sin\(\Theta\)

\(\left ( \frac{sin27^{\circ}}{cos63^{\circ}} \right )^{2}\)\(\left ( \frac{cos63^{\circ}}{sin27^{\circ}} \right )^{2}\) = \(\left (\frac{sin(90^{\circ}-63^{\circ})}{cos63^{\circ}}\right )^{2}\)\(\left (\frac{cos(90^{\circ}-27^{\circ})}{sin27^{\circ}}\right )^{2}\)

=\(\left ( \frac{cos63^{\circ}}{cos63^{\circ}} \right )^{2}\)\(\left ( \frac{sin27^{\circ}}{sin27^{\circ}} \right )^{2}\) =1-1 =0

So value of \(\left ( \frac{sin27^{\circ}}{cos63^{\circ}} \right )^{2}\)\(\left ( \frac{cos63^{\circ}}{sin27^{\circ}} \right )^{2}\) is 0

 

(v)\(\frac{tan35^{\circ}}{cot55^{\circ}}+\frac{cot78^{\circ}}{tan12^{\circ}}-1\)

 

Soln.(v)

We have to find:

\(\frac{tan35^{\circ}}{cot55^{\circ}}+\frac{cot78^{\circ}}{tan12^{\circ}}-1\)

Since tan (\(90^{\circ}-\Theta\))= cot\(\Theta\) and cot (\(90^{\circ}-\Theta\))=tan\(\Theta\) =1

So value of \(\frac{tan35^{\circ}}{cot55^{\circ}}+\frac{cot78^{\circ}}{tan12^{\circ}} is 1\)

 

(vi) \(\frac{sec70^{\circ}}{cosec20^{\circ}}+\frac{sin59^{\circ}}{cos31^{\circ}}\)

 

Soln.(vi)

We have to find: \(\frac{sec70^{\circ}}{cosec20^{\circ}}+\frac{sin59^{\circ}}{cos31^{\circ}}\)

Since \(\frac{sec70^{\circ}}{cosec20^{\circ}}+\frac{sin59^{\circ}}{cos31^{\circ}}\)and sec(\(90^{\circ}-\Theta\))=cosec\(\Theta\)

So

\(\frac{sec70^{\circ}}{cosec20^{\circ}}+\frac{sin59^{\circ}}{cos31^{\circ}}\)=\(\left (\frac{sec(90^{\circ}-20^{\circ})}{cosec20^{\circ}}\right )^{2}\)\(\left (\frac{sin(90^{\circ}-31^{\circ})}{cos31^{\circ}}\right )^{2}\)

=\(\frac{cosec20^{\circ}}{cosec20^{\circ}}+\frac{cos31^{\circ}}{cos31^{\circ}}\) =1+1 =2

So value of \(\frac{sec70^{\circ}}{cosec20^{\circ}}+\frac{sin59^{\circ}}{cos31^{\circ}}\) is 2

 

(vii) \(cosec31^{\circ}-sec59^{\circ}\).

 

Soln(vii)

We have to find: \(cosec31^{\circ}-sec59^{\circ}\)

Since \(cosec(90^{\circ}-\Theta )\)=sec\(\Theta\).So

= \(cosec31^{\circ}-sec59^{\circ}\)

= \(cosec\left ( 90^{\circ} -59^{\circ}\right )-sec59\) =\(sec59^{\circ}-sec59^{\circ}\) =0

So value of \(cosec31^{\circ}-sec59^{\circ}\) is 0

 

(viii)\(\left ( sin72^{\circ}+cos18^{\circ} \right )\) \(\left ( sin72^{\circ}-cos18^{\circ} \right )\)

 

Soln.(viii)

We have to find:  \(\left ( sin72^{\circ}+cos18^{\circ} \right )\) \(\left ( sin72^{\circ}-cos18^{\circ} \right )\)

Since \(sin(90^{\circ}-\Theta )\)=cos\(\Theta\),So

\(\left ( sin72^{\circ}+cos18^{\circ} \right )\) \(\left ( sin72^{\circ}-cos18^{\circ} \right )\)= \(\left ( sin72^{\circ} \right )^{2}\)\(\left ( cos18^{\circ} \right )^{2}\)

=\(\left [ sin\left ( 90^{\circ}-18^{\circ} \right ) \right ]^{2}-\left ( cos18^{\circ} \right )^{2}\)

=\(\left ( cos18^{\circ} \right )^{2}\)\(\left ( cos18^{\circ} \right )^{2}\)

=\(cos^{2}18^{\circ}-cos^{2}18^{\circ}\) =0

So value of \(\left ( sin72^{\circ}+cos18^{\circ} \right )\) \(\left ( sin72^{\circ}-cos18^{\circ} \right )\) is 0.

 

(ix)\(sin35^{\circ}sin55^{\circ}\)\(cos35^{\circ}cos55^{\circ}\)

 

Soln(ix)

We find :

\(sin35^{\circ}sin55^{\circ}\)\(cos35^{\circ}cos55^{\circ}\)

Since \(sin(90^{\circ}-\Theta )\)=cos\(\Theta\) and \(cos(90^{\circ}-\Theta )\)=sin\(\Theta\)

\(sin35^{\circ}sin55^{\circ}\)\(cos35^{\circ}cos55^{\circ}\)= \(sin(90^{\circ}-55^{\circ})sin55^{\circ}\)\(cos(90^{\circ}-55^{\circ})cos55^{\circ}\) =1-1 =0

So value of \(sin35^{\circ}sin55^{\circ}\)\(cos35^{\circ}cos55^{\circ}\) is  0

 

(x) \(tan48^{\circ}tan23^{\circ}tan42^{\circ}tan67^{\circ}\)

 

Soln.(x)

We have to find \(tan48^{\circ}tan23^{\circ}tan42^{\circ}tan67^{\circ}\)

Since \(tan(90^{\circ}-\Theta )\)=cot\(\Theta\). So

\(tan48^{\circ}tan23^{\circ}tan42^{\circ}tan67^{\circ}\)= \(tan\left ( 90^{\circ}-42^{\circ} \right )tan\left ( 90^{\circ}-67^{\circ} \right )tan42^{\circ}tan67^{\circ}\)

=\(cot42^{\circ}cot67^{\circ}tan42^{\circ}tan67^{\circ}\)

=\(\left ( tan67^{\circ}cot67^{\circ} \right )\left ( tan42^{\circ}cot42^{\circ} \right )\) =1×1 =1

So value of \(tan48^{\circ}tan23^{\circ}tan42^{\circ}tan67^{\circ}\) is 1

 

(xi)\(sec50^{\circ}sin40^{\circ}+cos40^{\circ}cosec50^{\circ}\)

 

Soln.(xi)

We have to find \(sec50^{\circ}sin40^{\circ}+cos40^{\circ}cosec50^{\circ}\)

Since \(cos(90^{\circ}-\Theta )\)=sin\(\Theta\),\(sec(90^{\circ}-\Theta )\)=cosec\(\Theta\) and sin\(\Theta\).cosec\(\Theta\)=1. So

\(sec50^{\circ}sin40^{\circ}+cos40^{\circ}cosec50^{\circ}\) = \(sec(90^{\circ}-40^{\circ})sin40^{\circ}\) +\(cos(90^{\circ}-50^{\circ})cosec50^{\circ}\) =1+1 =2

So value of \(sec50^{\circ}sin40^{\circ}+cos40^{\circ}cosec50^{\circ}\) is 2.

 

Q.3) Express cos\(75^{\circ}\)+cot\(75^{\circ}\) in terms of angle between 00 and 300.

 

Soln. 3 :

Given that: cos\(75^{\circ}\)+cot\(75^{\circ}\)

=cos\(75^{\circ}\)+cot\(75^{\circ}\)

= \(cos\left ( 90^{\circ}-15^{\circ} \right )+ cot\left ( 90^{\circ}-15^{\circ} \right )\)

=sin \(15^{\circ}\)+tan \(15^{\circ}\)

Hence the correct answer is sin \(15^{\circ}\)+tan \(15^{\circ}\)

 

Q.4) If  sin3A = cos(A – 260), where 3A is an acute angle, find the value of A.

 

Soln.4:

We are given 3A is an acute angle

We have: sin3A=cos(A-\(26^{\circ}\))

\(\Rightarrow\)sin3A=sin(\(90^{\circ}\)-(A-\(26^{\circ}\)))

\(\Rightarrow\) sin3A=sin(\(116^{\circ}\)-A)

\(\Rightarrow\) 3A=\(116^{\circ}\)-A

\(\Rightarrow\) 4A=\(116^{\circ}\)

\(\Rightarrow\)A=\(29^{\circ}\)

Hence thecorrect answer is \(29^{\circ}\)

 

Q.5)If A, B, C are the interior angles of a triangle ABC, prove that,

(i) \(tan\left ( \frac{C+A}{2} \right )=cot\frac{B}{2}\)

(ii) \(sin\left ( \frac{B+C}{2} \right )=cos\frac{A}{2}\)

 

Soln.5:

(i)We have to prove: \(tan\left ( \frac{C+A}{2} \right )=cot\frac{B}{2}\)

Since we know that in triangle ABC

A+B+C=180

\(\Rightarrow\)C+A=\(180^{\circ}\)-B

\(\Rightarrow\) \(\frac{C+A}{2}=90^{\circ}-\frac{B}{2}\)

\(\Rightarrow\)\(tan\frac{C+A}{2}=tan\left ( 90^{\circ} \frac{B}{2}\right )\)

\(\Rightarrow\)\(tan\left ( \frac{C+A}{2} \right )=cot\frac{B}{2}\)

Hence proved

 

(ii)We have to prove : \(sin\left ( \frac{B+C}{2} \right )=cos\frac{A}{2}\)

Since we know that in triangle ABC

A+B+C=180

\(\Rightarrow\)B+C=\(180^{\circ}\)-A

\(\Rightarrow\) \(\frac{B+C}{2}=90^{\circ}-\frac{A}{2}\)

\(\Rightarrow\)\(sin\frac{B+C}{2}=sin\left ( 90^{\circ} \frac{A}{2}\right )\)

\(sin\left ( \frac{B+C}{2} \right )=cos\frac{A}{2}\)

Hence proved

 

Q.6)Prove that :

(i)tan\(20^{\circ}\)tan\(35^{\circ}\)tan\(45^{\circ}\)tan\(55^{\circ}\)tan\(70^{\circ}\) = 1

(ii)sin\(48^{\circ}\).sec\(48^{\circ}\)+cos\(48^{\circ}\).cosec\(42^{\circ}\) = 2

(iii) \(\frac{sin70^{\circ}}{cos20^{\circ}}+\frac{cosec20^{\circ}}{sec70^{\circ}}-2cos70^{\circ}.cosec20^{\circ}=0\)

(iv) \(\frac{cos80^{\circ}}{sin10^{\circ}}+cos59^{\circ}.cosec31^{\circ}=2\)

 

Soln.6:

(i)Therefore

tan\(20^{\circ}\)tan\(35^{\circ}\)tan\(45^{\circ}\)tan\(55^{\circ}\)tan\(70^{\circ}\)

=\(tan\left ( 90^{\circ} – 70^{\circ}\right )\) \(tan\left ( 90^{\circ} – 55^{\circ}\right )\) tan\(45^{\circ}\)tan\(55^{\circ}\)tan\(70^{\circ}\)

=cot\(70^{\circ}\)cot\(55^{\circ}\)tan\(45^{\circ}\)tan\(55^{\circ}\)tan\(70^{\circ}\)

= \(\left ( tan70^{\circ} cot70^{\circ} \right )\left ( tan55^{\circ} cot55^{\circ}\right )tan45^{\circ}\) =1x1x1 =1

Hence proved

 

(ii) We will simplify the left hand side

sin\(48^{\circ}\).sec\(48^{\circ}\)+cos\(48^{\circ}\).cosec\(42^{\circ}\)=sin\(48^{\circ}\). \(sec\left ( 90^{\circ}-48^{\circ} \right )\)cos\(48^{\circ}\).\(cosec\left ( 90^{\circ}-48^{\circ} \right )\)

=sin\(48^{\circ}\).cos\(48^{\circ}\)+cos\(48^{\circ}\).sin\(48^{\circ}\) =1+1 =2

Hence proved

 

(iii) We have,\(\frac{sin70^{\circ}}{cos20^{\circ}}+\frac{cosec20^{\circ}}{sec70^{\circ}}-2cos70^{\circ}.cosec20^{\circ}=0\)

So we will calculate left hand side

\(\frac{sin70^{\circ}}{cos20^{\circ}}+\frac{cosec20^{\circ}}{sec70^{\circ}}-2cos70^{\circ}.cosec20^{\circ}=0\)= \(\frac{sin70^{\circ}}{cos20^{\circ}}+\frac{cos70^{\circ}}{sin20^{\circ}}-2cos70^{\circ}.cosec(90^{\circ}-70^{\circ})\)

=\(\frac{sin(90^{\circ}-20^{\circ})}{cos20^{\circ}}\)+ \(\frac{cos(90^{\circ}-20^{\circ})}{sin20^{\circ}}\)\(2cos70^{\circ}.cosec(90^{\circ}-70^{\circ})\)

=\(\frac{cos20^{\circ}}{cos20^{\circ}}+\frac{sin20^{\circ}}{sin20^{\circ}}-2×1\) =1+1-2 =2-2 =0

Hence proved

 

(iv)We have \(\frac{cos80^{\circ}}{sin10^{\circ}}+cos59^{\circ}.cosec31^{\circ}=2\)

We will simplify the left hand side

\(\frac{cos80^{\circ}}{sin10^{\circ}}+cos59^{\circ}.cosec31^{\circ}\)= \(\frac{cos\left ( 90^{\circ}-10^{\circ} \right )}{sin10^{\circ}}+cos59^{\circ}.cosec\left ( 90^{\circ}-59^{\circ} \right )\)

=\(\frac{sin10^{\circ}}{sin10^{\circ}}+cos59^{\circ}.sec59^{\circ}\) =1+1 =2

Hence proved.

 

Question 7

If A,B,C are the interior of triangle ABC , show that

(i) \(\sin (\frac{B+C}{2})=\cos \frac{A}{2}\)

 

Solution

A+B+C=1800

B + C = 1800\(\frac{A}{2}\)

LHS=RHS

 

(ii)\(\cos (90^{0}-\frac{A}{2})=\sin \frac{A}{2}\)

LHS=RHS

 

 

Question 8

If \(2\Theta +45^{0}\,and\,30-\Theta\) are acute angles , find the degree measure of

 \(\Theta\) satisfying \(\sin (20+45^{0})=\cos (30^{0}+\Theta )\)

 

Solution

Here 20+450=\(\sin (60^{0}+\Theta )\)

 

We know that ,(\((90^{0}-\Theta )\)= \(\cos (\Theta )\)

= \(\sin (2\Theta +45^{0})=\sin (90^{0}-(30^{0}-\Theta ))\)

= \(\sin (2\Theta +45^{0})=\sin (90^{0}-30^{0}-\Theta )\)

= \(\sin (2\Theta +45^{0})=\sin (60^{0}+\Theta )\)

On equating sin of angle of we get,

\(2\Theta +45^{0}=60^{0}+\Theta\) = \(\Theta=15^{0}\)

 

Question 9

If \(\Theta\) is appositive acute angle such that \(\sec \Theta=\csc 60^{0}\) , find

\(2\cos ^{2}\Theta -1\)

 

Solution

We know that, \(\sec (90^{0}-\Theta )=\csc ^{2}\Theta\)

= \(\sec (\Theta )=\sec (90^{0}-60^{0})\)

= \(\Theta=30^{0}\) = \(2\cos ^{2}\Theta -1\)

= \(2\cos ^{2}30-1\)

= \(2(\frac{\sqrt{3}}{2})^{2}-1\)

= \(2(\frac{3}{4})-1\) =\((\frac{3}{2})-1\) = \((\frac{1}{2})\)

 

Q10.If \(\sin 3\Theta = \cos \left ( \Theta -6^{\circ} \right )\;where\;3\Theta\;and \;\Theta -6^{circ}\) acute angles, find the value of \(\Theta\).

 

Soln:     

We have, \(\sin 3\Theta = \cos \left ( \Theta -6^{\circ} \right )\)

\(\cos \left ( 90^{\circ}+3\Theta \right ) = \cos \left ( \Theta -6^{\circ} \right )\)

\(90^{\circ} – 3\Theta = \Theta – 6^{\circ}\)

\(-3\Theta -\Theta = 6^{\circ}-90^{\circ}\)

\(-4\Theta = 96^{\circ}\)

\(\Theta = \frac{-96^{\circ}}{-4}= 24^{\circ} \)

 

Q11.If \(\sec 2A = \csc (A-42^{\circ})\) where 2A  is acute angle, find the value of A.

 

Soln:we know that \(\sec \left(90- 3Theta \right) = \csc \Theta\)

\(\sec 2A = \sec \left ( 90 – (A -42) \right )\)

\(\sec 2A = \sec \left ( 90 – A + 42) \right ) \)

\(\sec 2A = \sec \left ( 132 – A) \right ) \)

Now equating both the angles we get

2A = 132 – A

A = \( = \frac{132}{3} \)

A = 44