# 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-

$\dpi{100} \\sine \:of\: \angle A=\frac{side \:opposite \:to \:angle \:A}{hypotenuse}=\frac{BC}{AC}\\\\cosine \:of\: \angle A=\frac{side \:adjacent\: to \:angle \:A}{hypotenuse}=\frac{AB}{AC}\\\\tangent \:of\: \angle A=\frac{side \:opposite \:to \:angle \:A}{side \:adjacent\: to \:angle \:A}=\frac{BC}{AB}\\\\cosecant \:of\:\angle A=\frac{hypotenuse}{side \:opposite \:to \:angle \:A}=\frac{AC}{BC}\\\\secant \:of \:\angle A= \frac{hypotenuse}{side \:adjacent\: to \:angle \:A}=\frac{AC}{AB}\\\\cotangent \:of\: \angle A=\frac{side \:adjacent\: to \:angle \:A}{side \:opposite \:to \:angle \:A}=\frac{AB}{BC}$

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

 $\dpi{100} \angle A$ $\dpi{100} 0^o$ $\dpi{100} 30^o$ $\dpi{100} 45^o$ $\dpi{100} 60^o$ $\dpi{100} 90^o$ Sin A 0 $\dpi{100} \frac{1}{2}$ $\dpi{100} \frac{1}{\sqrt{2}}$ $\dpi{100} \frac{\sqrt{3}}{2}$ 1 Cos A 1 $\dpi{100} \frac{\sqrt{3}}{2}$ $\dpi{100} \frac{1}{\sqrt{2}}$ $\dpi{100} \frac{1}{2}$ 0 Tan A 0 $\dpi{100} \frac{1}{\sqrt{3}}$ 1 $\dpi{100} \sqrt{3}$ Not defined Cosec A Not defined 2 $\dpi{100} \sqrt{2}$ $\dpi{100} \frac{2}{\sqrt{3}}$ 1 Sec A 1 $\dpi{100} \frac{2}{\sqrt{3}}$ $\dpi{100} \sqrt{2}$ 2 Not defined Cot A Not defined $\dpi{100} \sqrt{3}$ 1 $\dpi{100} \frac{1}{\sqrt{3}}$ 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$

We have,
In $\Delta \: ABC$$\angle$B = 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 $\angle$C
So,  $\sin C = P/H = BA/AC = 24/25$
and  $\cos C = B/H = BC/AC = 7/25$

We have, $\Delta$PQR 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

Suppose $\Delta$ABC is a right-angled triangle in which $\angle B = 90^0$ and we have $\dpi{100} \sin A=\frac{3}{4},$
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 $\tan A = \frac{BC}{AB} = \frac{3}{\sqrt{7}}$

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                 $\sec A = \frac{AC}{AB} = \frac{17}{8}$

We have,
$\dpi{100} \sec \theta =\frac{13}{12},$

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 $\angle$B 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,
$\sin \theta = \frac{BC}{AC}=\frac{5}{13}$
$\cos \theta = \frac{BA}{AC}=\frac{12}{13}$

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

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

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

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

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, $\angle$A = $\angle$ B   [angle opposite to equal sides are equal]

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       $\cos \theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}$

$\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}$
$\dpi{100} (i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta$
$=(\frac{7}{8})^2 = \frac{49}{64}$

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

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

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

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,

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

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

We have,
PR + QR = 25 cm.............(i)
PQ = 5 cm
and $\angle Q =90^0$
According to question,
In triangle $\Delta$PQR,
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,
$\\\sin P= \frac{QR}{PR}= 12/13\\ \cos P = \frac{PQ}{RP} = 5/13\\ \therefore \tan P = \frac{\sin P}{\cos P} = 12/5$

(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 $\Theta .$

(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

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

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

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

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 :

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

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

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

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

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

$\frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$..................(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;
$\\\Rightarrow \frac{1/2+1-2/\sqrt{3}}{2/\sqrt{3}+1/2+1}\\ \Rightarrow\frac{3/2-2/\sqrt{3}}{3/2+2/\sqrt{3}}\\ \Rightarrow \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}\\ \Rightarrow \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}\\ \Rightarrow \frac{43-24\sqrt{3}}{11}$

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

$\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}$.....................(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;

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

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

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

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)

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

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

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

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

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

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

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

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

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 $\dpi{100} \tan (A-B)= \frac{1}{\sqrt{3}};$  $0^{o} B,$ find $A \: and \: B.$

Given that,
$\tan (A+B) = \sqrt{3} =\tan 60^0$
So, $\dpi{100} A + B = 60^o$..........(i)
$\tan (A-B) = 1/\sqrt{3} =\tan 30^0$
therefore, $\dpi{100} A - B = 30^o$.......(ii)
By solving the equation (i) and (ii) we get;

$\dpi{100} A = 45^o$ and $\dpi{100} B = 15^o$

$(i) \sin (A + B) = \sin A + \sin B$
$(ii)$  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 $A=0^{o}$

(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

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

$\frac{\sin 18^{o}}{\cos 72^{o}}$
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, $\frac{\cos 72^0}{\cos 72^0} = 1$

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

$\frac{\tan 26^{o}}{\cot 64^{o}}$
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,
$\cot64^o/\cot 64^o = 1$

$(iii) \cos 48^{o}-\sin 42^{o}$

$\cos 48^{o}-\sin 42^{o}$
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,
$\sin42^{o}-\sin 42^{o} = 0$

$(iv) cosec \: 31^{o}-\sec 59^{o}$

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

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

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

$\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$
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.

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

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

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

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

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.

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

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

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

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.

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

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) $\tan A = \frac{1}{\cot A}$

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

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

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

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

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

$\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}$....................(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 $\sin^2\theta +\cos^2\theta = 1$)

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

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

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

Therefore,

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

$(i) 9\sec^{2}A-9\tan^{2}A=$

(A) 1    (B) 9    (C) 8    (D) 0

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, $9\times 1 = 9$

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

(A) 0    (B) 1    (C) 2    (D) –1

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

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

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

The correct option is (D)

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

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

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

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,

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

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

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.

$(ii) \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A$

We need to prove-

$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A$

taking LHS;
$\\=\frac{\cos^2A+1+\sin^2A+2\sin A}{\cos A(1+\sin A)}\\\\ =\frac{2(1+\sin A)}{\cos A(1+\sin A)}\\\\ =2/\cos A = 2\sec A$

= RHS

Hence proved.

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

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

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.

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

[Hint : Simplify LHS and RHS separately]

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 $\csc ^{2}A= 1+\cot ^{2}A$

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.

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

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.

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.

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

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

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

[Hint : Simplify LHS and RHS separately]

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.

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

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

## 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!

Exams
Articles
Questions