NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Edited By Komal Miglani | Updated on Mar 31, 2025 07:45 PM IST | #CBSE Class 10th

Have you ever wondered how engineers design huge buildings, how sailors and pilots navigate through their journey, or how shadows change their length throughout the day? All of these answers can be found in Trigonometry, a fascinating branch of mathematics. From the latest NCERT syllabus for class 10, the chapter introduction to trigonometry contains the basic concepts of trigonometry, like trigonometric ratios, trigonometric identities, trigonometric ratios of some specific angles, and trigonometric ratios of complementary angles. Understanding these concepts will make students more efficient in solving problems involving height, distance, and angles. This chapter will also build a strong foundation for advanced trigonometric concepts, which have many practical real-life applications like construction, navigation, architecture, etc.

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  1. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
  2. Introduction to Trigonometry Class 10 NCERT Solutions - Important Formulae
  3. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Exercises)
  4. Trigonometry Class 10 Solutions - Exercise Wise
  5. NCERT Solutions for Class 10 Maths: Chapter Wise
  6. Importance of Solving NCERT Questions of Class 10 Maths Chapter 8
  7. NCERT Solutions of Class 10 - Subject Wise
  8. NCERT Exemplar Solutions - Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

This article on NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry offers clear and step-by-step solutions for the exercise problems in the NCERT Class 10 Maths Book. Students who are in need of introduction to trigonometry class 10 solutions will find this article very useful. It covers all the important class 10 maths chapter 8 question answers of introduction to trigonometry. These class 10 Introduction to Trigonometry ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 10 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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Introduction to Trigonometry Class 10 NCERT Solutions - Important Formulae

Arc Length in a Circle:

  • If an arc of length $l$ in a circle with radius $r$ subtends an angle of $\theta$ radians.

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Arc length and radius of a circle

$l = r\theta$ and $θ = \frac{l}{r}$
Conversion Between Radian and Degree Measures:

  • Radian Measure = ($\frac{π}{180}$) × Degree Measure
  • Degree Measure = ($\frac{180}{π}$) × Radian Measure
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Trigonometric Ratios for Right Triangles:

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

Trigonometric ratios

$\begin{aligned} & \text { sine of } \angle A=\frac{\text { side oppositeto angle } A}{\text { hypotenuse }}=\frac{B C}{A C} \\ & \text { cosine of } \angle A=\frac{\text { side adjacent to angle } A}{\text { hypotenuse }}=\frac{A B}{A C} \\ & \text { tangent of } \angle A=\frac{\text { side opposite to angle } A}{\text { side adjacentto angle } A}=\frac{B C}{A B} \\ & \text { cosecant of } \angle A=\frac{\text { hypotenuse }}{\text { side opposite to angle } A}=\frac{A C}{B C} \\ & \text { secant of } \angle A=\frac{\text { hypotenuse }}{\text { side adjacent to angle } A}=\frac{A C}{A B} \\ & \text { cotangent of } \angle A=\frac{\text { side adjacent to angle } A}{\text { side opposite to angle } A}=\frac{A B}{B C}\end{aligned}$

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

$\angle A$30°45°60°90°

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


Also, in a right triangle with an angle $\theta$

Trigonometric ratios of right angle triangle

  • sin θ = Opposite/Hypotenuse

  • cos θ = Adjacent/Hypotenuse

  • tan θ = Opposite/Adjacent

  • cosec θ = Hypotenuse/Opposite

  • sec θ = Hypotenuse/Adjacent

  • cot θ = Adjacent/Opposite

Reciprocal Trigonometric Ratios:

  • sin θ = 1/(cosec θ)

  • cosec θ = 1/(sin θ)

  • cos θ = 1/(sec θ)

  • sec θ = 1/(cos θ)

  • tan θ = 1/(cot θ)

  • cot θ = 1/(tan θ)

Trigonometric Ratios of Complementary Angles:

For an angle 'θ', the trigonometric ratios of its complementary angle (90° – θ) are:

  • sin (90° – θ) = cos θ

  • cos (90° – θ) = sin θ

  • tan (90° – θ) = cot θ

  • cot (90° – θ) = tan θ

  • sec (90° – θ) = cosec θ

  • cosec (90° – θ) = sec θ

Trigonometric Identities:

  • sin2 θ + cos2 θ = 1

  • sin2 θ = 1 – cos2 θ

  • cos2 θ = 1 – sin2 θ

  • cosec2 θ – cot2 θ = 1

  • cot2 θ = cosec2 θ – 1

  • sec2 θ – tan2 θ = 1

  • tan2 θ = sec2 θ – 1

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Exercises)

NCERT introduction to trigonometry class 10 questions and answers
Exercise: 8.1, Page number: 121, Total questions: 11

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:

ABC right angle triangle

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,
$AC^2 = AB^2 + BC^2$
⇒ $AC = \sqrt{AB^2+BC^2}$
⇒ $AC = \sqrt{576+49}$
⇒ $AC = \sqrt{625}$
⇒ $AC = 25$ cm
Now,
(i) $\sin A = \frac{P}{H} = \frac{BC}{AC} = \frac{7}{25}$
$\cos A = \frac{B}{H} = \frac{BA}{AC} = \frac{24}{25}$

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

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

PQR right angle triangle

Answer:
We have $\Delta$PQR
According to Pythagoras, in a right-angled triangle, the lengths 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 the question,
$\tan P -\cot R$ = $\frac{RQ}{QP}-\frac{QR}{PQ} = \frac{5}{12} - \frac{5}{12} = 0$

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

Answer:

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

ABC right angle triangle

Let the length of AB be 4 units and the length of BC = 3 units. 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,
$15 \cot A=8$ $\Rightarrow \cot A =\frac{8}{15}$
It implies that in the triangle ABC, in which $\angle B =90^0$. The length of AB is 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,
$\sec \theta =\frac{13}{12},$

This means that the hypotenuse of the triangle is 13 units, and the base is 12 units.
Let ABC be a right-angled triangle in which $\angle$ B is 90 and AB is the base, BC is the perpendicular height, and AC is the hypotenuse.

ABC right angle triangle

By using Pythagoras' theorem,
$BC = \sqrt{169-144}=\sqrt{25}$
$\Rightarrow BC = 5$ units

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 $\cos A =\cos B$.

ABC right angle triangle

According to the question, in triangle ABC,
$\cos A =\cos B$
$\frac{AC}{AB}=\frac{BC}{AB}$
$\Rightarrow AC = AB$
Therefore, $\angle$ A = $\angle$ B [angles 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 (BC) = 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, $\sin \theta = \frac{AB}{AC} = \frac{8}{\sqrt{113}}$
and $\cos \theta = \frac{BC}{AC} = \frac{7}{\sqrt{113}}$

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

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

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 the length of the perpendicular is 3 units

ABC right angle triangle

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

So,
$\tan A = \frac{BC}{AB} = \frac{3}{4}$
$\cos A = \frac{AB}{AC} = \frac{4}{5}$
$\sin A = \frac{BC}{AC} = \frac{3}{5}$
$\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A$
Putting the values of the above trigonometric ratios, we get;
$\Rightarrow \frac{1-\frac{9}{16}}{1+\frac{9}{16}} = \frac{16}{25}-\frac{9}{25}$
$\Rightarrow \frac{7}{25} = \frac{7}{25}$
LHS $=$ 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 $\tan A =\frac{1}{\sqrt{3}}$ $\Rightarrow A=30^0$

ABC right angle triangle

According to question, $\tan A =\frac{1}{\sqrt{3}} = \frac{BC}{AB}$
By using Pythagoras' theorem,
$\\AC^2 = AB^2+BC^2\Rightarrow AC = \sqrt{1+3} =\sqrt{4}$
$\Rightarrow AC = 2$
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,

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

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


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:

PQR right angle triangle

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

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

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

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 the $\cos A$ 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 introduction to trigonometry class 10 questions and answers
Exercise: 8.2, Page number: 127, Total questions: 4

Q1. Evaluate the following :

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

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

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 the question,

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

Q1. Evaluate the following :

$(iii)\: \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

Answer:
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
We know the value of
$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ , $\sec 30^{\circ}= \frac{2}{\sqrt {3}}$ and $\operatorname{cosec}30^{\circ} =2$ ,
After putting these values
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{(2+2\sqrt{3})}{ \sqrt{3}}}$
$\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}$
$\\=\frac{2\sqrt{6}-2\sqrt{18}}{-16}\\ =2(\frac{\sqrt{6}-3\sqrt{3}}{-16}) = \frac{3\sqrt{3}-\sqrt{6}}{8}$

Q1. Evaluate the following :

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

Answer:

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

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,
$\\\cos 60^0 = \frac{1}{2}= \sin 30^0, \sec 30^0 = \frac{2}{\sqrt{3}}, \tan 45^0 = 1, \cos 30^0 = \frac{\sqrt{3}}{2}$
By substituting all these values in equation(i), we get;

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

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

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:


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

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 $tan 30^{o}={\frac{1}{\sqrt{3}}}$

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

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)
$\tan (A-B) = \frac{1}{\sqrt{3}} =\tan 30^0$
therefore, $A - B = 30^o$ .......(ii)
By solving equations (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, $\\\sin(45^0+45^0) = \sin 45^0+\sin 45^0$
$\sin 90^0 = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \neq 1$

(ii) True,
Take $\theta = 0^0,\ 30^0,\ 45^0$
when
$\theta$ = 0 then zero(0),
$\theta$ = 30 then value of $\sin \theta$ is $\frac{1}{2}$ = 0.5
$\theta$ = 45 then value of $\sin \theta$ is $\frac{1}{\sqrt2}$ = 0.707

(iii) False,
$\cos 0^0 = 1,\ \cos 30^0 = \frac{\sqrt{3}}{2}= 0.87,\ \cos 45^0 = \frac{1}{\sqrt2}= 0.707$

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

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


NCERT introduction to trigonometry class 10 questions and answers

Exercise: 8.3, Page number: 131, Total questions: 4

Q1. Express the trigonometric ratios $\sin A,\sec, and \tan A$ 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
$\sec ^2 A-\tan ^2 A=1$
$\Rightarrow \sec^2 A=1+\frac{1}{\cot^2\theta}$
$\Rightarrow \sec A=\frac{\sqrt{\cot^2\theta+1}}{\cot\theta}$

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

Q3. 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 sec2A - tan2A = 1

Therefore, equation (i) becomes, $9\times 1 = 9$

Q3. 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;
$=(1+\frac{\sin \theta}{\cos \theta} +\frac{1}{\cos \theta})(1+\frac{\cos\theta}{\sin \theta} -\frac{1}{sin\theta})$
$= \frac{(1+\sin\theta+\cos\theta)}{\cos\theta.\sin\theta}\times (\frac{(\sin\theta+\cos\theta-1}{\sin\theta.\cos\theta})$
$= \frac{(\sin\theta+\cos\theta)^2-1^2}{\sin\theta.\cos\theta}$
$= \frac{\sin^2\theta+\cos^2\theta+2\sin\theta.\\cos\theta-1}{\sin\theta.\cos\theta}$
$= 2\times\frac{\sin\theta.\cos\theta}{\sin\theta.\cos\theta}$
$= 2$

Q3. 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)

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

Q3. 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)

$\frac{1+\tan ^{2}A}{1+\cot ^{2}A}$

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

therefore,

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

Q4. 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.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

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

Answer:

We need to prove-

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

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

= RHS

Hence proved.

Q4. 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;

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

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

Q4. 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;

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

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

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

LHS = RHS

Hence proved.

Q4. 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.

Q4. 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.

Q4. 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 $\cos2\theta = 2\cos^2\theta-1=\cos^2\theta-\sin^2\theta$ ]

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

Hence proved.

Q4. 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 the equation,
$(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A$ ..................(i)

Taking LHS;

$(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}$
$\\\Rightarrow \sin^2 A+\csc^2A +2+\cos^2A+\sec^2A+2= 1+2+2+(1+\cot^2A)+(1+\tan^2A)$
[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

Q4. 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;
$\\\Rightarrow (cosec A-\frac{1}{\csc A})(\sec A-\frac{1}{\sec A})=\frac{(cosec^2-1)}{cosec A}\times\frac{\sec^2A-1}{\sec A}=\frac{\cot^2A}{cosec A}.\frac{\tan^2A}{\sec A}=\sin A .\cos A$

Taking RHS;

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

LHS = RHS

Hence proved.

Q4. 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;

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

LHS = RHS

Hence proved.

Trigonometry Class 10 Solutions - Exercise Wise

Here are the exercise-wise links for the NCERT class 10 chapter 8, Introduction to Trigonometry:

NCERT Solutions for Class 10 Maths: Chapter Wise

Here, students can find chapter-wise solutions.

Importance of Solving NCERT Questions of Class 10 Maths Chapter 8

  • Solving these NCERT questions will help students understand the basic concepts of trigonometry easily.
  • Students can practice various types of questions, which will improve their problem-solving skills.
  • These NCERT exercises cover all the important topics and concepts so that students can be well-prepared for various exams.
  • By solving these NCERT trigonometry problems, students will get to know about all the real-life applications of trigonometry.

NCERT Solutions of Class 10 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 10:

NCERT Exemplar Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 10 NCERT:

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 10:

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Solutions for Class 10 Maths Chapter 8?

This NCERT Solutions for Class 10 Maths Chapter 8 contains the basic concepts of trigonometry like trigonometric ratios, trigonometric identities, and their applications in real world.

2. How to solve trigonometric identities in Class 10 Maths Chapter 8?

To solve trigonometric identities in Class 10 Maths Chapter 8, use the basic trigonometric identities as sin2 θ + cos2 θ = 1 and try to express terms in sine and cosine. Also, use simplification and factorisation to solve these types of problems.

3. What is the easiest way to understand trigonometric ratios?

The easiest way to understand trigonometric ratios is by taking a right-angle triangle where sine, cosine, and tangent represent ratios of different sides as 
sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse
tan θ = Opposite / Adjacent

4. How can I score full marks in Class 10 Trigonometry?

To score full marks in Class 10 Trigonometry, memorise all the important trigonometric ratios, identities, and formulas. Practice NCERT examples, exercises, and questions from the previous years. Also, regularly revise and give mock tests to improve scores.

5. What are the real-life applications of trigonometry?

Trigonometry is used in various fields of real life. It is mainly used for calculating the height and distance of real-life objects, and for building bridges and roads. Also, it is used in navigation systems like GPS.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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