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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Edited By Apoorva Singh | Updated on Sep 06, 2023 07:46 PM IST | #CBSE Class 10th

Introduction To Trigonometry Class 10 NCERT Solution

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry explains the relation between the angles and sides of a right angle triangle. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry are a valuable resource for students as they assist in both understanding the concepts and performing well on the CBSE Class 10 board examination. Introduction to trigonometry class 10 solutions are created by subject experts and include answers for all questions in the textbook. They are also updated to align with the latest CBSE Syllabus for 2022-23 and exam pattern.

This Story also Contains
  1. Introduction To Trigonometry Class 10 NCERT Solution
  2. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
  3. Introduction to Trigonometry Class 10 NCERT Solutions - Important Formuale
  4. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Intext Questions and Exercise)
  5. Features of Trigonometry Class 10 NCERT Solutions
  6. Trigonometry Class 10 Solutions - Exercise Wise
  7. NCERT Solutions for Class 10 Maths - Chapter 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

In addition to providing a strong foundation for the concepts in this chapter, the NCERT Books class 10 trigonometry solutions also allow students to clear their doubts and grasp the fundamentals. They also provide helpful guidance for solving challenging problems in each exercise of Chapter 8 Introduction to Trigonometry.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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Introduction to Trigonometry Class 10 NCERT Solutions - Important Formuale

Arc Length in a Circle:

  • If an arc of length 'l' in a circle with radius 'r' subtends an angle of 'θ' radians.

1694008793688

l = rθ

θ = l/r

Conversion Between Radian and Degree Measures:

  • Radian Measure = (π/180) × Degree Measure

  • Degree Measure = (180/π) × Radian Measure

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Apply

Trigonometric Ratios for Right Triangles:

In a right triangle with an angle 'θ':

1694008794138

  • 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

Free download NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Intext Questions and Exercise)

Trigonometry chapter class 10 ncert solutions 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:

1635935005859 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

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

1635935041656

Answer:


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

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,
1635935058122

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 \angle B is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.

1635935082533 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

sdfgjhkjhdcgv 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]

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
1635935097081

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

FireShot%20Capture%20151%20-%20Careers360%20CMS%20-%20cms-articles%20-%20learn 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:

sdfghjkdfghj 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

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 ]


Trigonometry chapter class 10 ncert solutions 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)


Trigonometry chapter class 10 ncert solutions 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)


Introduction to trigonometry class 10 solutions 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
1635934455463

(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 sec2A-tan2A=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.

(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)}\\\\ =2/\cos A = 2\sec A

= 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 a 3 - b 3 =(a - b) (a 2 + b 2 +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.

Features of Trigonometry Class 10 NCERT Solutions

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 NCERT solutions for class 10 maths chapter 8. These NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry are designed to provide assistance for homework and for preparing the board examinations.

Trigonometry Class 10 Solutions - Exercise Wise

Trigonometry Class 10 Topic-

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

sdfgdfghjkl

\\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 Wise

Chapter No.

Chapter Name

Chapter 1

Real Numbers

Chapter 2

Polynomials

Chapter 3

Pair of Linear Equations in Two Variables

Chapter 4

Quadratic Equations

Chapter 5

Arithmetic Progressions

Chapter 6

Triangles

Chapter 7

Coordinate Geometry

Chapter 8

Introduction to Trigonometry

Chapter 9

Applications of Trigonometry

Chapter 10

Circles

Chapter 11

Constructions

Chapter 12

Areas Related to Circles

Chapter 13

Surface Areas and Volumes

Chapter 14

Statistics

Chapter 15

Probability

Benefits of NCERT Solutions for Class 10 Maths Chapter 8

  • These Class 10 Maths Chapter 8 NCERT solutions are prepared by the experts. Hence these solutions are 100 per cent reliable.

  • The Trigonometry Class 10 will be beneficial for Class 10 board exams and for higher studies as well.

  • NCERT chapter 8 Maths Class 10 solutions will help in building the basic concepts of trigonometry and bring forth some easy ways to solve the questions.

NCERT Solutions of Class 10 - Subject Wise

How to use NCERT solutions for Class 10 Maths chapter 8 Introduction to Trigonometry?

  • Firstly, learn all the concepts given in the NCERT book. Memorise all the trigonometric ratios, angle values, and trigonometric identities.

  • Now practice exercises by referring to the NCERT Class 10 Maths solutions chapter 8.

  • As the NCERT Solutions for Class 10 Maths Chapter 8 PDF Download is not available. So you can save the webpage to practice the solutions offline.

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

NCERT Exemplar solutions - Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. Whether this unit Introduction to Trigonometry is helpful for higher studies?

Trigonometry is a most important field in mathematics which is useful in almost every field including architecture, electronics, seismology, meteorology, oceanography etc. Trigonometry problems can be solved using NCERT book and NCERT exemplar for Class 10 Mathematics. Students can download trigonometry class 10 NCERT solutions pdf for ease and study both online and offline mode.

2. How many chapters are there in the Class 10 Maths?

There are a total of 15 chapters in the Class 10 Maths NCERT syllabus. Questions from all the exercises of each chapter are available in the Careers 360 website. The chapter wise link provided navigate you the solution page of the respective chapter. Students can download NCERT solutions for class 10 maths chapter 8 pdf using the link give above in this article.

3. List out the frequently-asked topics of class 10 maths trigonometry Solutions in the CBSE exam of Class 10 Maths.

The topics that are commonly covered in CBSE Maths exams for introduction to trigonometry class 10 maths chapter 8 solutions include: Introduction to Trigonometry, Trigonometric Identities, Trigonometric Ratios of Specific Angles, Trigonometric Ratios of Complementary Angles, and Trigonometric Identities. CBSE board class 10 paper is entirely based on the NCERT.

4. Why should we download NCERT Solutions for maths class 10 chapter 8 from Careers360?

Careers360 offers precise answers to the questions found in the NCERT Solutions for class 10 chapter 8 maths. These solutions can be accessed online and also downloaded in PDF format. The solutions for maths chapter 8 class 10 are explained by experts in a clear and concise manner, and diagrams are included as needed.

5. Is NCERT Solutions for ncert class 10 trigonometry important from the exam point of view?

Yes, All chapters in the introduction to trigonometry class 10 solutions are essential for both board exams and future grades. It's crucial for students to practise all the questions in NCERT Solutions for Class 10 Maths Chapter 8 in order to achieve high marks on the exams.

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Read Complete Answer
SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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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sadafreen356 03 July,2024
can we apply for diploma if we have passed 10th cbse lo

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.

You can pursue diplomas at various institutions like:


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Popular CBSE Class 10th Questions

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

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0.16\; J
Option 3)

1.00\; J

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0.67\; J
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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 :

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2.45×10−3 kg

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 6.45×10−3 kg
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In the reaction,

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

Option 1)

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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

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

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3.125 × 10-2
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1.25 × 10-2

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2.5 × 10-2
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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

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increase two fold
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remain unchanged

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With increase of temperature, which of these changes?

Option 1)

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Fraction of solute present in water

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

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twice that in 60 g carbon

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6.023 × 1022
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558.5 × 6.023 × 1023
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less than 3

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more than 3 but less than 6
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more than 6 but less than 9

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As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

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