NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

 

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry - The chapter trigonometry has been generated from the right angle triangle. The relation between the angles and sides of a right angle triangle is called trigonometry. In solutions of NCERT class 10 maths chapter 8 Introduction to Trigonometry, there is a detailed explanation for each question of practice exercise. In this particular chapter, you will study some ratios of a right-angled triangle with respect to its acute angles, and this ratio is called a trigonometric ratio of the angles. CBSE NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry is also covering the questions based on trigonometric ratios. Unit 5 "Trigonometry" holds 12 marks out of 80 marks in the maths paper of CBSE board examination and we can expect 2-3 questions from this chapter of total around 8 marks. There is a total of 4 exercises with 27 questions in the entire chapter. NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry are designed to provide assistance for homework and for preparing the board examinations. If you want to download the NCERT solutions for free then you can have it by clicking on the link.

The trigonometric ratios of the angle A in right triangle ABC are defined as follows-

The trigonometric ratios

\\sine \:of\: \angle A=\frac{side \:opposite \:to \:angle \:A}{hypotenuse}=\frac{BC}{AC}\\\\cosine \:of\: \angle A=\frac{side \:adjacent\: to \:angle \:A}{hypotenuse}=\frac{AB}{AC}\\\\tangent \:of\: \angle A=\frac{side \:opposite \:to \:angle \:A}{side \:adjacent\: to \:angle \:A}=\frac{BC}{AB}\\\\cosecant \:of\:\angle A=\frac{hypotenuse}{side \:opposite \:to \:angle \:A}=\frac{AC}{BC}\\\\secant \:of \:\angle A= \frac{hypotenuse}{side \:adjacent\: to \:angle \:A}=\frac{AC}{AB}\\\\cotangent \:of\: \angle A=\frac{side \:adjacent\: to \:angle \:A}{side \:opposite \:to \:angle \:A}=\frac{AB}{BC}

The values of all the trigonometric ratios of 0°, 30°, 45°, 60°, and 90° are-

   \angle A

    0^o

   30^o   

   45^o

   60^o

  90^o

Sin A

  0

   \frac{1}{2}

  \frac{1}{\sqrt{2}}

  \frac{\sqrt{3}}{2}  

1

Cos A

1

  \frac{\sqrt{3}}{2}

  \frac{1}{\sqrt{2}}    

   \frac{1}{2}    

0

Tan A

 0

  \frac{1}{\sqrt{3}}

1

  \sqrt{3}

 Not defined

Cosec A

 Not defined

 2

\sqrt{2}

  \frac{2}{\sqrt{3}}

1

Sec A  

  1

  \frac{2}{\sqrt{3}}

  \sqrt{2}

   2

 Not defined

Cot A

 Not defined

  \sqrt{3} 

  1

  \frac{1}{\sqrt{3}}

 0

 

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In \Delta \: ABC, right-angled at B, AB = 24 \: cm, BC = 7 \: cm. Determine : (i)\; \sin A, \cos A(ii)\; \sin C, \cos C

Answer:


We have,
In \Delta \: ABC\angleB = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm 
So, by using Pythagoras theorem,
\\AC^2 = AB^2 + BC^2\\ AC = \sqrt{AB^2+BC^2}
Therefore, AC = \sqrt{576+49}
                   AC = \sqrt{625}
                    AC = 25 cm

Now,
(i) \sin A = P/H = BC/AB = 7/25
    \cos A = B/H = BA/AC = 24/25

(ii)  For angle C, AB is perpendicular to the base (BC).  Here B  indicates to Base and P means perpendicular wrt angle \angleC
So,  \sin C = P/H = BA/AC = 24/25       
and  \cos C = B/H = BC/AC = 7/25

Q2 In Fig. 8.13, find \tan P - \cot R.

            

Answer:


We have, \DeltaPQR is a right-angled triangle, length of PQ and PR are 12 cm and 13 cm respectively.
So, by using Pythagoras theorem,
QR = \sqrt{13^2-12^2}
QR = \sqrt{169-144}
QR = \sqrt{25} = 5\ cm

Now, According to question,
\tan P -\cot R = \frac{RQ}{QP}-\frac{QR}{PQ}
                               = 5/12 - 5/12  = 0

Q3 If \sin A=\frac{3}{4}, calculate \cos A and \tan A.

Answer:

Suppose \DeltaABC is a right-angled triangle in which \angle B = 90^0 and we have \sin A=\frac{3}{4},
So, 

Let the length of AB be 4 unit and the length of BC = 3 unit
So, by using Pythagoras theorem,
AB = \sqrt{16-9} = \sqrt{7} units
Therefore, 
\cos A = \frac{AB}{AC} = \frac{\sqrt{7}}{4}  and \tan A = \frac{BC}{AB} = \frac{3}{\sqrt{7}}

Q4 Given 15 \: \cot A=8, find \sin A and \sec A.

Answer:

We have,
15 \: \cot A=8, \Rightarrow \cot A =8/15 
It implies that In the triangle ABC in which \angle B =90^0. The length of AB be 8 units and the length of BC  = 15 units

Now, by using Pythagoras theorem,
AC = \sqrt{64 +225} = \sqrt{289}
\Rightarrow AC =17 units

So,                  \sin A = \frac{BC}{AC} = \frac{15}{17}
and                 \sec A = \frac{AC}{AB} = \frac{17}{8} 

Q5 Given \sec \theta =\frac{13}{12}, calculate all other trigonometric ratios.

Answer:

We have,
\sec \theta =\frac{13}{12},

It means the Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a right-angled triangle in which \angleB is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.


By using Pythagoras theorem,

BC = \sqrt{169-144}=\sqrt{25}
BC = 5 unit

Therefore,
\sin \theta = \frac{BC}{AC}=\frac{5}{13} 
\cos \theta = \frac{BA}{AC}=\frac{12}{13}

\tan \theta = \frac{BC}{AB}=\frac{5}{12}

\cot \theta = \frac{BA}{BC}=\frac{12}{5}

\sec \theta = \frac{AC}{AB}=\frac{13}{12}

\csc \theta = \frac{AC}{BC}=\frac{13}{5}

Q6 If \angle A and \angle B are acute angles such that \cos A = \cos B, then show that \angle A =\angle B.

Answer:

We have, A and B are two acute angles of triangle ABC and \cos A =\cos B


According to question,
In triangle ABC,
\cos A =\cos B
 \frac{AC}{AB}=\frac{BC}{AB}
\Rightarrow AC = AB
Therefore, \angleA = \angle B   [angle opposite to equal sides are equal]

Q7 If \cot \theta =\frac{7}{8}, evaluate:(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} (ii)\; \cot ^{2}\theta

Answer:

Given that,
\cot \theta =\frac{7}{8}
\therefore perpendicular  (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which \angle B =90^0
Now, By using Pythagoras theorem,
AC^2 = AB^2+BC^2
AC = \sqrt{64 +49} =\sqrt{113}

So,        \sin \theta = \frac{BC}{AC} = \frac{8}{\sqrt{113}}       
and       \cos \theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}

\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}
(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta
 =(\frac{7}{8})^2 = \frac{49}{64}

(ii)\; \cot ^{2}\theta 
  =(\frac{7}{8})^2 = \frac{49}{64}

Q8 If 3\cot A=4, check wether \frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A or not.

Answer:

Given that,
3\cot A=4, 
\Rightarrow \cot = \frac{4}{3} = \frac{base}{perp.}
ABC is a right-angled triangle in which \angle B =90^0 and the length of the base AB is 4 units and length of perpendicular is 3 units

By using Pythagoras theorem,
In triangle ABC,
\\AC^2=AB^2+BC^2\\ AC = \sqrt{16+9}\\ AC = \sqrt{25}
AC = 5 units

So, 
\tan A = \frac{BC}{AB} = \frac{3}{4}
\cos A = \frac{AB}{AC} = \frac{4}{5}
\sin A = \frac{BC}{AC} = \frac{3}{5}
 \frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A
Put the values of above trigonometric ratios, we get;
\Rightarrow \frac{1-9/4}{1+9/4} = \frac{16}{25}-\frac{9}{25}
\Rightarrow -\frac{5}{13} \neq \frac{7}{25}
LHS \neq RHS

Q9 In triangle ABC, right-angled at B, if  \tan A =\frac{1}{\sqrt{3}},find the value of:

(i) \sin A\: \cos C + \cos A\: \sin C
(ii) \cos A\: \cos C + \sin A\: \sin C

Answer:

Given a triangle ABC, right-angled at  B and \tan A =\frac{1}{\sqrt{3}}   \Rightarrow A=30^0

According to question,
\tan A =\frac{1}{\sqrt{3}} = \frac{BC}{AB}
By using Pythagoras theorem,
\\AC^2 = AB^2+BC^2\\ AC = \sqrt{1+3} =\sqrt{4}
AC = 2 
Now, 
\\\sin A = \frac{BC}{AC} = \frac{1}{2}\\ \sin C =\frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos C = \frac{BC}{AC} = \frac{1}{2}

Therefore,
  
(i) \sin A\: \cos C + \cos A\: \sin C
\\\Rightarrow \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\\ \Rightarrow1/4 +3/4\\ \Rightarrow4/4 = 1

(ii) \cos A\: \cos C + \sin A\: \sin C
\\\Rightarrow \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}\\ \Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}

 

Q10 In \Delta \: PQR, right-angled at Q, PR + QR = 25\: cm and PQ = 5 \: cm. Determine the values of  \sin P, \cos P \: and\: \tan P.

Answer:


We have,
PR + QR = 25 cm.............(i)
PQ = 5 cm
and \angle Q =90^0
According to question,
In triangle \DeltaPQR,
By using Pythagoras theorem,
\\PR^2 = PQ^2+QR^2\\ PQ^2 =PR^2-QR^2 \\ 5^2= (PR-QR)(PR+QR)\\ 25 = 25(PR-QR) \\
PR - QR = 1........(ii)

From equation(i) and equation(ii), we get;
PR = 13 cm  and QR = 12 cm.

therefore,
\\\sin P= \frac{QR}{PR}= 12/13\\ \cos P = \frac{PQ}{RP} = 5/13\\ \therefore \tan P = \frac{\sin P}{\cos P} = 12/5

Q11 State whether the following are true or false. Justify your answer.

              (i) The value of \tan A is always less than 1.
              (ii) \sec A=\frac{12}{5} for some value of angle A.
              (iii) \cos A is the abbreviation used for the cosecant of angle A.
              (iv) \cot A is the product of cot and A.
              (v) \sin \Theta =\frac{4}{3}  for some angle \Theta .

Answer:

(i) False,
because \tan 60 = \sqrt{3}, which is greater than 1

(ii) TRue,
because \sec A \geq 1

(iii) False, 
Because \cos A abbreviation is used for cosine A.

(iv) False,
because the term \cot A is a single term, not a product.

(v) False,
because \sin \theta lies between (-1 to +1) [-1\leq \sin \theta\leq 1]

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.2

Q1 Evaluate the following :

      (i) \sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}

Answer:

\sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}
As we know,
the value of \sin 60^0 = \sqrt{3}/2 = \cos 30^0 , \sin 30^0 = 1/2=\cos 60^0 
\\\Rightarrow \frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+ \frac{1}{2}.\frac{1}{2}\\
  =\frac{3}{4}+\frac{1}{4}
 =1
    

Q1 Evaluate the following :

     (ii)\: 2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}

Answer:

We know the value of

 \tan 45^0 = 1 and

 \cos 30^0 = \sin 60^0 = \frac{\sqrt{3}}{2}
According to question,

=2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}
\\=2(1)^2+ (\frac{\sqrt{3}}{2})^2-(\frac{\sqrt{3}}{2})\\=2

Q1 Evaluate the following :

 (iii)\: \frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}

Answer:

\frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}
we know the value of

 \cos 45 = 1/\sqrt{2} , \sec 30^0 = 2/\sqrt {3} and cosec \:30 =2,

After putting these values 
=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}
=\frac{1/\sqrt{3}}{(2+2\sqrt{3})/ \sqrt{3}}
\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}
\\=\frac{2\sqrt{6}-2\sqrt{18}}{-16}\\ =2\frac{\sqrt{6}-3\sqrt{3}}{-16} = \frac{3\sqrt{3}-\sqrt{6}}{8}

Q1 Evaluate the following :

      (iv)\: \frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}

Answer:

\frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}..................(i)
It is known that the values of the given trigonometric functions,
\\\sin 30^0 = 1/2=cos 60^0\\ \tan 45^0 = 1=\cot 45^0\\ \sec 30^0 = 2/\sqrt{3}=cosec 60^0\\
Put all these values in equation (i), we get;
\\\Rightarrow \frac{1/2+1-2/\sqrt{3}}{2/\sqrt{3}+1/2+1}\\ \Rightarrow\frac{3/2-2/\sqrt{3}}{3/2+2/\sqrt{3}}\\ \Rightarrow \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}\\ \Rightarrow \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}\\ \Rightarrow \frac{43-24\sqrt{3}}{11}

Q1 Evaluate the following :

        (v)\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}

Answer:

\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}.....................(i)
We know the values of-
\\\cos 60^0 = 1/2= \sin 30^0\\ \sec 30^0 = 2/\sqrt{3}\\ \tan 45^0 = 1\\ \cos 30^0 = \sqrt{3}/2
By substituting all these values in equation(i), we get;

\\\Rightarrow \frac{5.(\frac{1}{2})^2+4.(\frac{2}{\sqrt{3}}) ^2-1}{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}\\ \Rightarrow \frac{5/4-1+16/3}{1}\\ \Rightarrow \frac{1/4+16/3}{1}\\ \Rightarrow\frac{67}{12}

Q2 Choose the correct option and justify your choice :

      (i)\, \frac{2\: \tan 30^{o}}{1+\tan ^{2}30^{o}}=

   (A)\: \sin 60^{o}                (B)\: \cos 60^{o}             (C)\: \tan 60^{o}            (D)\: \sin 30^{o}

Answer:

Put the value of tan 30 in the given question-
\frac{2\: \tan 30^{o}}{1+\tan ^{2}30^{o}}=\frac{2\times1/\sqrt{3}}{1+(1/\sqrt{3})^2}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\sqrt{3}/2 = \sin 60^0

The correct option is (A)

Q2 Choose the correct option and justify your choice :

  (ii)\: \frac{1-\tan^{2}45^{o}}{1+ \tan^{2}45^{o}}=

   (A)\: \tan \: 90^{o}               (B)\: 1            (C) \: \sin 45^{o}             (D) \: 0

Answer:

The correct option is (D)
\frac{1-\tan^{2}45^{o}}{1+ \tan^{2}45^{o}}=
We know that \tan 45 = 1
So, \frac{1-1}{1+1}=0

Q2 Choose the correct option and justify your choice :

 (iii)\sin \: 2A=2\: \sin A is true when A =

 (A)0^{o}                (B)\: 30^{o}                (C)\: 45^{o}                (D)\: 60^{o}

Answer:

The correct option is (A)
\sin \: 2A=2\: \sin A
We know that \sin 2A = 2\sin A \cos A
So, 2\sin A \cos A = 2\sin A
\\\cos A = 1\\ A = 0^0

Q2 Choose the correct option and justify your choice :

(iv)\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=

  (A)\: \cos 60^{o}                (B)\: \sin 60^{o}              (C)\: \tan 60^{o}            (D)\: \sin 30^{o}  

Answer:

Put the value of tan 30^{o}={\1/\sqrt{3}}

\\=\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=\frac{2\times 1/\sqrt{3}}{1-(\frac{1}{\sqrt{3}})^2}\\ =\frac{2/\sqrt{3}}{1-1/3}\\ =\frac{2}{\sqrt{3}}\times \frac{3}{2}\\ =\sqrt{3}= \tan 60^0

The correct option is (C)

Q3 If  \tan (A+B)=\sqrt{3} and \tan (A-B)= \frac{1}{\sqrt{3}};  0^{o}<A+B\leq 90^{o};A> B, find A \: and \: B.

Answer:

Given that,
\tan (A+B) = \sqrt{3} =\tan 60^0
So, A + B = 60^o..........(i)
\tan (A-B) = 1/\sqrt{3} =\tan 30^0
therefore, A - B = 30^o.......(ii)
By solving the equation (i) and (ii) we get;

A = 45^o and B = 15^o

Q4 State whether the following are true or false. Justify your answer.

             (i) \sin (A + B) = \sin A + \sin B
             (ii)  The value of \sin \theta  increases as \theta increases.
             (iii) The value of \cos \theta increases as \theta increases.
             (iv)\sin \theta =\cos \theta  for all values of \theta .
             (v) \cot A is not defined for A=0^{o}

Answer:

(i) False,
Let A = B =45^0
Then, \\\sin(45^0+45^0) = \sin 45^0+\sin 45^0\\ \sin 90^ = 1/\sqrt{2}+q/\sqrt{2}\\ 1 \neq \sqrt{2}

 

(ii) True,
Take \theta = 0^0,\ 30^0,\ 45^0
whent 
\theta = 0 then zero(0),
 \theta = 30 then value of \sin \theta is 1/2 = 0.5
\theta = 45 then value of \sin \theta is 0.707

 

(iii) False,
\cos 0^0 = 1,\ \cos 30^0 = \sqrt{3}/2= 0.87,\ \cos 45^0 = 1\sqrt{2}= 0.707

 

(iv) False,
Let \theta = 0 
\\\sin 0^0 = \cos 0^0\\ 0 \neq 1

 

(v) True,
\cot 0^0 = \frac{\cos 0^0}{\sin 0^0}=\frac{1}{0}  (not defined)

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

Q1 Evaluate :

   (i)  \frac{\sin 18^{o}}{\cos 72^{o}}

Answer:

\frac{\sin 18^{o}}{\cos 72^{o}}
We can write the above equation as; 
=\frac{\sin (90^0-72^0)}{\cos 72^0}
By using the identity of  \sin (90^o-\theta) = \cos \theta
Therefore, \frac{\cos 72^0}{\cos 72^0} = 1

So, the answer is 1.

Q1 Evaluate :

    (ii) \frac{\tan 26^{o}}{\cot 64^{o}}

Answer:

\frac{\tan 26^{o}}{\cot 64^{o}}
The above equation can be written as ;

\tan (90^o-64^o)/\cot 64^o.........(i)
It is known that, \tan (90^o-\theta) = \cot \theta
Therefore, equation (i) becomes,
\cot64^o/\cot 64^o = 1

So, the answer is 1. 

Q1 Evaluate :

  (iii) \cos 48^{o}-\sin 42^{o}

Answer:

\cos 48^{o}-\sin 42^{o}
The above equation can be written as ;
\cos (90^o-42^{o})-\sin 42^{o}....................(i)
It is known that \cos (90^o-\theta) = \sin \theta
Therefore, equation (i) becomes,
\sin42^{o}-\sin 42^{o} = 0

So, the answer is 0.

Q1 Evaluate :

 (iv) cosec \: 31^{o}-\sec 59^{o}

Answer:

cosec \: 31^{o}-\sec 59^{o}

This equation can be written as;
cosec 31^o - \sec(90^o-31^o).................(i)
We know that  \sec(90^o-\theta) = cosec \theta

Therefore, equation (i) becomes;
cosec 31^o - cosec\ 31^o = 0

So, the answer is 0. 

Q2 Show that :

 (i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1

Answer:

\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1
Taking Left Hand Side (LHS)
\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}
\Rightarrow \tan 48^{o}\tan 23^{o}\tan (90^o-48^{o})\tan (90^o-23^{o})
\Rightarrow \tan 48^{o}\tan 23^{o}\cot 48^{o}\cot23^{o}     [it is known that \tan (90^0-\theta = \cot\theta) and \cot\theta\times \tan \theta =1 
=1

Hence proved.

Q2 Show that :

   (ii) \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Answer:

\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Taking Left Hand Side (LHS)
=\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}
=\cos 38^{o}\cos (90^o-38^{o})-\sin 38^{o}\sin (90^o-38^{o})
=\cos 38^{o}\sin38^{o}-\sin 38^{o}\cos 38^{o} [it is known that \sin(90^0-\theta) =\cos \theta and \cos(90^0-\theta) =\sin \theta ]
= 0

Q3 If  \tan 2A= \cot (A-18^{o}), where 2A is an acute angle, find the value of A.

Answer:

We have,
 \tan 2A  = \cot (A - \18^{0}
we know that, \\\cot (90^0-\theta) = \tan\theta
\\\Rightarrow \cot (90^0-2A) = \cot (A-18^0)\\\Rightarrow 90^0-2A = A - 18^0\\\Rightarrow 3A = 108\\\Rightarrow A = 108/3 = 36^0

Q4 If \tan A= \cot B, prove that A+B= 90^{o}.

Answer:

We have,
\tan A= \cot B
and we know that \tan (90^0 - \theta)= \cot \theta
therefore, 
\tan A= \tan(90^0- B)
A = 90 - B 
A + B = 90
Hence proved.

Q5 If \sec 4A= cosec (A-20^{o}), where 4A is an acute angle, find the value of A.

Answer:

We have,
\sec 4A= cosec (A-20^{o}), Here 4A is an acute angle
According to question,
We know that  cosec(90^0-\theta)= \sec \theta
cosec(90^0-4A)= cosec (A-20^{o})

\\\Rightarrow 90 - 4A = A - 20\\\\ \Rightarrow 5A=110\\\\ \Rightarrow A=\frac{110}{5}\\\:\\ \Rightarrow A=22^o

Q6 If A,B and C are interior angles of a triangle ABC, then show that

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

Answer:

Given that,
A, B and C are interior angles of \Delta ABC 
To prove - \sin (\frac{B+C}{2})= \cos \frac{A}{2}

Now,
In triangle \Delta ABC
       A + B + C  = 180^0
\Rightarrow B + C = 180 - A
\Rightarrow B + C/2 = 90^0 - A/2
\sin \frac{B+C}{2}=\sin (90^0-A/2)
\sin \frac{B+C}{2}=\cos A/2
Hence proved.

Q7 Express sin 67^{o}+\cos 75^{o}in terms of trigonometric ratios of angles between 0^{o} and 45^{o}.

Answer:

By using the identity of \sin\theta and \cos\theta
sin 67^{o}+\cos 75^{o}
We know that, 
\sin(90-\theta) =\cos \theta and \cos(90-\theta) =\sin \theta
the above equation can be written as;
=\sin (90^0-23^0)+\cos(90^0-15^0)
=\sin (15^0)+\cos(23^0)

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

Q1 Express the trigonometric ratios \sin A,\sec A and \tan A in terms of cot A.

Answer:

We know that \csc^2A -\cot^2A = 1
(i)
\\\Rightarrow \frac{1}{\sin^2A}= 1+\cot^2A\\ \Rightarrow\sin^2A = \frac{1}{1+\cot^2A}\\ \Rightarrow \sin A = \frac{1}{\sqrt{1+\cot^2A}}

(ii) We know the identity of 
\\\sec^2A - \tan^2A = 1\\ \frac{1}{\cos^2A} = 1+\tan^2A=1+\frac{1}{\cot ^2A}\\ \frac{\cot^2A}{1+\cot^2A}=\cos^2A\\ \frac{\cot A}{\sqrt{1+\cot�}} = \cos A

(iii) \tan A = \frac{1}{\cot A}

Q2 Write all the other trigonometric ratios of \angle A  in terms of \sec A.

Answer:

We know that the identity \sin^2 A + \cos^2 =1
\\\sin^2 A =1- \cos^2 \\ sin^2A = 1-\frac{1}{\sec^2A}
               =\frac{\sec^2A -1}{\sec^2A}
\sin A =\sqrt{\frac{\sec^2A -1}{\sec^2A}}
            =\frac{{}\sqrt{\sec^2A -1}}{\sec A}

cosec A =\frac{\sec A}{\sqrt{\sec^2A -1}}

\tan A = \frac{\sin A}{\cos A} = \sqrt{\sec^2A -1}

\cot A =\frac{1}{ \sqrt{\sec^2A -1}}

Q3 Evaluate :

     (i)\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}

Answer:

\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}....................(i)

The above equation can be written as;

\\=\frac{\sin ^{2}63^{o}+\sin ^{2}(90^0-63^{o})}{\cos ^{2}(90^0-73^{o})+\cos ^{2}73^{o}}\\\\ =\frac{\sin ^{2}63^{o}+\cos ^{2}63^{o}}{\sin ^{2}73^{o}+\cos ^{2}73^{o}}\\\\ = 1 
(Since \sin^2\theta +\cos^2\theta = 1)

Q3 Evaluate :

   (ii)\sin 25^{o}\cos 65^{o}+\cos 25^{o}\sin 65^{o}

Answer:

\sin 25^{o}\cos 65^{o}+\cos 25^{o}\sin 65^{o}

We know that 
\\\sin(90^0-\theta) = \cos \theta \\\cos (90^0-\theta) = \sin \theta

Therefore,

\\\sin 25^{o}\cos (90^0-25^{o})+\cos 25^{o}\sin (90^0-25^{o})\\ \sin 25^0.\sin 25^0 + \cos 25^0.\cos 25^0\\ sin^2 25^0+\cos ^225^0\\ 1

Q4 Choose the correct option. Justify your choice.

     (i) 9\sec^{2}A-9\tan^{2}A=

    (A) 1    (B) 9    (C) 8    (D) 0

Answer:

The correct option is (B) = 9

9\sec^2A-9 \tan ^2A = 9(\sec^2A- \tan ^2A).............(i)

and it is known that \sec^2\theta -\tan^2\thea = 1
Therefore, equation (i) becomes, 9\times 1 = 9

Q4 Choose the correct option. Justify your choice.

      (ii)(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\: \theta )=

       (A) 0    (B) 1    (C) 2    (D) –1

Answer:

The correct option is (C)

(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\: \theta ).......................(i)

we can write his above equation as;
\\=(1+\sin \theta/\cos \theta +1/\cos \theta )(1+\cos\theta/\sin \theta -1/sin\theta )\\\\= \frac{(1+\sin\theta+\cos\theta)}{\cos\theta.\sin\theta}\times (\frac{(\sin\theta+\cos\theta-1}{\sin\theta.\cos\theta})\\\\= \frac{(\sin\theta+\cos\theta)^2-1^2}{\sin\theta.\cos\theta}\\\\= \frac{\sin^2\theta+\cos^2\theta+2\sin\theta.\\cos\theta-1}{\sin\theta.\cos\theta}\\\\= 2\times\frac{\sin\theta.\cos\theta}{\sin\theta.\cos\theta}
= 2

Q4 Choose the correct option. Justify your choice.

      (iii) (\sec A+\tan A)(1-\sin A)=

      (A)\sec A   (B)\sin A  (C)cosec A   (D) \cos A

Answer:

The correct option is (D)

(\sec A+\tan A)(1-\sin A)=
\\ \Rightarrow ( \frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A)\\\\ \Rightarrow \frac{1+\sin A}{\cos A}(1-\sin A)\\\\ \Rightarrow \frac{1-\sin^2 A}{\cos A}\\\\\Rightarrow \cos A

Q4 Choose the correct option. Justify your choice.

     (iv) \frac{1+\tan ^{2}A}{1+\cot ^{2}A}=

     (A) \sec ^{2}A   (B) -1  (C) \cot ^{2}A    (D) \tan ^{2}A

Answer:

The correct option is (D)

\frac{1+\tan ^{2}A}{1+\cot ^{2}A}..........................eq (i)

The above equation can be written as;

We know that \cot A = \frac{1}{\tan A}

therefore,

\\\Rightarrow \frac{1+\tan ^{2}A}{1+\frac{1}{\tan ^{2}}A}\\ \Rightarrow \tan^2A\times(\frac{1+\tan^2 A}{1+\tan^2A})\\ \Rightarrow \tan^2A

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (i) (\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }

Answer:

We need to prove-
(\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }

Now, taking LHS,

(\csc\theta-\cot\theta)^2 = (\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta})^2
                                  \\= (\frac{1-\cos\theta}{\sin\theta})^2\\ =\frac{(1-\cos\theta)(1-\cos\theta)}{\sin^2\theta}\\
                                   \\=\frac{(1-\cos\theta)(1-\cos\theta)}{1-\cos^2\theta}\\\\ =\frac{(1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\\\ =\frac{1-\cos\theta}{1+\cos\theta}

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (iii)\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta

           [Hint : Write the expression in terms of \sin \theta and \cos\theta]
 

Answer:

We need to prove-
\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \:cosec \theta

Taking LHS;

\\\Rightarrow \frac{\tan^2 \theta }{\tan \theta-1 }+\frac{1}{\tan\theta(1-\tan \theta) }\\\\\ \Rightarrow\frac{\tan^3\theta-\tan^4\theta+\tan\theta-1}{(\tan\theta-1).\tan\theta.(1-\tan\theta)}\\\\ \Rightarrow \frac{(\tan^3\theta-1)(1-\tan\theta)}{\tan\theta.(\tan\theta-1)(1-\tan\theta)}\\
By using the identity a3 - b3 =(a - b) (a2 + b2+ab)

\\\Rightarrow \frac{(\tan\theta -1)(\tan^2\theta+1+\tan\theta)}{\tan\theta(\tan\theta -1a)}\\\\ \Rightarrow \tan\theta+1+\frac{1}{\tan\theta}\\\\ \Rightarrow 1+\frac{1+\tan^2\theta}{\tan\theta}\\\\ \Rightarrow 1+\sec^2\theta \times \frac{1}{\tan\theta}\\\\ \Rightarrow 1+\sec\theta.\csc\theta\\\\ =RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (iv)\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}

          [Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-
\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}

taking LHS;

\\\Rightarrow \frac{1+\sec A}{\sec A}\\ \Rightarrow (1+\frac{1}{\cos A})/\sec A\\ \Rightarrow 1+\cos A

Taking RHS;
We know that identity 1-\cos^2\theta = \sin^2\theta

\\\Rightarrow \frac{\sin ^{2}A}{1-\cos A}\\ \Rightarrow \frac{1-\cos^2 A}{1-\cos A}\\ \Rightarrow \frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}\\ \Rightarrow 1+\cos A

LHS = RHS 

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (v) \frac{\cos A-\sin A+1}{\cos A+\sin A-1}= \csc A+\cot A , using the identity \csc ^{2}A= 1+\cot ^{2}A

Answer:

We need to prove -
\frac{\cos A-\sin A+1}{\cos A+\sin A-1}= cosec A+\cot A

Dividing the numerator and denominator by \sin A, we get;

\\=\frac{\cot A-1+\csc A}{\cot A +1-\csc A}\\\\= \frac{(\cot A+\csc A)-(\csc^2 A-\cot^2A)}{\cot A +1-\csc A}\\\\= \frac{(\csc A+\cot A)(1-\csc A+\cot A)}{\cot A +1-\csc A}\\\\= \csc A+\cot A\\\\ =RHS

Hence Proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

   (vi)\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A

Answer:

We need to prove -
\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A
Taking LHS;
By rationalising the denominator, we get;

\\= \sqrt{\frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}\\\\ = \sqrt{\frac{(1+\sin A)^2}{1-\sin^2A}}\\\\ =\frac{1+\sin A}{\cos A}\\\\ = \sec A + \tan A\\\\ = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(vii)\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta

Answer:

We need to prove -
\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta

Taking LHS;
[we know the identity \cos2\theta = 2\cos^2\theta-1=\cos^2\theta-\sin^2\theta]

\\\Rightarrow \frac{\sin \theta(1 -2\sin ^{2}\theta) }{\cos\theta(2\cos ^{2}\theta -1) }\\\\ \Rightarrow \frac{\sin\theta(\sin^2\theta+\cos^2\theta-2\sin^\theta)}{\cos\theta.\cos2\theta}\\\\ \Rightarrow \frac{\sin\theta.\cos2\theta}{\cos\theta.\cos2\theta}\\\\ \Rightarrow \tan\theta =RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

    (viii)(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A

Answer:

Given equation,
(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A..................(i)

Taking LHS;

(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}
\\\Rightarrow \sin^2 A+\csc^2A +2+\cos^2A+\sec^2A+2\\\\ \Rightarrow 1+2+2+(1+\cot^2A)+(1+\tan^2A)
[since \sin^2\theta +\cos^2\theta = 1, \csc^2\theta-\cot^2\theta =1, \sec^2\theta-\tan^2\theta=1]

\\7+\csc^2A+\tan^2A\\ =RHS

Hence proved

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

   (ix)\:(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}

                      [Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-
(coesc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}
Taking LHS;
\\\Rightarrow (cosec A-\frac{1}{\csc A})(\sec A-\frac{1}{\sec A})\\\\ \Rightarrow\frac{(cosec^2-1)}{cosec A}\times\frac{\sec^2A-1}{\sec A}\\\\ \Rightarrow\frac{\cot^2A}{cosec A}.\frac{\tan^2A}{\sec A}\\\\ \Rightarrow\sin A .\cos A

Taking RHS;

\\\Rightarrow\frac{1}{\sin A/\cos A+\cos A/\sin A}\\\\ \Rightarrow\frac{\sin A .\cos A}{\sin^2A+\cos^2A}\\\\ \Rightarrow \sin A.\cos A

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

          (x) (\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A

Answer:

We need to prove,
(\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A

Taking LHS;

\\\Rightarrow \frac{1+\tan ^{2}A}{1+\cot ^{2}A} = \frac{\sec^2A}{\csc^2A}=\tan^2A

Taking RHS;

\\=(\frac{1-\tan A}{1-\cot A})^2\\\\= (\frac{1-\sin A/\cos A }{1-\cos A /\sin A})^2\\\\ = \frac{(\cos A -\sin A)^2(\sin^2A)}{(\sin A-\cos A)^2(\cos^2A)}\\\\ =\tan^2A

LHS = RHS

Hence proved.

NCERT solutions for class 10 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers

Chapter 2

NCERT solutions for class 10 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Chapter 4

CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations

Chapter 5

NCERT solutions for class 10 chapter 5 Arithmetic Progressions

Chapter 6

Solutions of NCERT class 10 maths chapter 6 Triangles

Chapter 7

CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry

Chapter 8

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Chapter 9

Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry

Chapter 10

CBSE NCERT solutions class 10 maths chapter 10 Circles

Chapter 11

NCERT solutions  for class 10 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles

Chapter 13

CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes

Chapter 14

NCERT solutions for class 10 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 10 maths chapter 15 Probability

Solutions of NCERT class 10 subject wise

NCERT solutions for class 10 maths

Solutions of NCERT for class 10 science

How to use NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry?

  • First of all, be familiar with the concepts of the right-angle triangle.

  • This chapter needs high memorization means you have to memorize the trigonometric ratios, angle values, and trigonometric identities.

  • Once you have done the basic part then you can jump to the practice exercises.

  • While doing the practice exercises if you stuck anywhere then you take the help of NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry.

  • After doing all these you can practice the last 5 years question papers of board examinations.

Keep working hard & happy learning!

 

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