Trigonometric Identities

XYZ is a right-angled triangle in which ∠XZY = 90° and ∠XYZ = θ, which in an acute angle.

10th Grade Trigonometric Identities

We know,

sin θ = \(\frac{\textrm{Opposite}}{\textrm{Hypotenuse}}\) = \(\frac{o}{h}\);

cos θ = \(\frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}\) = \(\frac{a}{h}\);

tan θ = \(\frac{\textrm{Opposite}}{\textrm{Adjacent}}\) = \(\frac{o}{a}\);

csc θ = \(\frac{\textrm{Hypotenuse}}{\textrm{Opposite}}\) = \(\frac{h}{o}\);

sec θ = \(\frac{\textrm{Hypotenuse}}{\textrm{Adjacent}}\) = \(\frac{h}{a}\);

cot θ = \(\frac{\textrm{Adjacent}}{\textrm{Opposite}}\) = \(\frac{a}{o}\).

Let’s multiply sin θ and csc θ

sin θ ∙ csc θ = \(\frac{o}{h}\) × \(\frac{h}{o}\) = 1

Therefore, csc θ = \(\frac{1}{sin θ}\)

Now, multiply cos θ and sec θ

cos θ ∙ sec θ = \(\frac{a}{h}\) × \(\frac{h}{a}\) = 1

Therefore, sec θ = \(\frac{1}{cos θ}\)

Again, multiply tan θ and cot θ

tan θ ∙ cot θ = \(\frac{o}{a}\) × \(\frac{a}{a}\) = 1

Therefore, cot θ = \(\frac{1}{tan θ}\)

Let’s divide sin θ by cos θ

sin θ ÷ cos θ = \(\frac{sin θ}{cos θ}\) = \(\frac{\frac{o}{h}}{\frac{a}{h}}\) = \(\frac{o}{a}\) = tan θ.

Therefore, tan θ = \(\frac{sin θ}{cos θ}\).

Similarly, divide cos θ by sin θ

cos θ ÷ sin θ = \(\frac{cos θ}{sin θ}\) = \(\frac{\frac{a}{h}}{\frac{o}{h}}\) = \(\frac{a}{o}\) = cot θ.

Therefore, cot θ = \(\frac{cos θ}{sin θ}\).

 

If a relation of equality between two expressions involving trigonometric ratios of an angle θ holds true for all values of θ then the equality is called a trigonometric identity. But it holds true only for some values of θ, the equality gives a trigonometric equation. There are a number of fundamental trigonometric identities.

I.  sin\(^{2}\) θ + cos\(^{2}\) θ = 1

We have, sin θ = \(\frac{o}{h}\) and cos θ = \(\frac{a}{h}\).

Therefore, sin\(^{2}\) θ + cos\(^{2}\) θ = \(\frac{o^{2}}{h^{2}}\) + \(\frac{a^{2}}{h^{2}}\).

                                       = \(\frac{o^{2} + a^{2}}{h^{2}}\)

                                       = \(\frac{h^{2}}{h^{2}}\); [Since, in the right-angled ∆XYZ, o\(^{2}\) + a\(^{2}\) = h\(^{2}\) (by Pythagoras’ Theorem)]

                                        = 1.

Therefore, sin\(^{2}\) θ + cos\(^{2}\) θ = 1.

Consequently, 1 - sin\(^{2}\) θ = cos\(^{2}\) θ and 1 - cos\(^{2}\) θ = sin\(^{2}\) θ.

 

II. sec\(^{2}\) θ - tan\(^{2}\) θ = 1

We have, sec θ = \(\frac{h}{a}\) and tan θ = \(\frac{o}{a}\).

Therefore, sec\(^{2}\) θ - tan\(^{2}\) θ = \(\frac{h^{2}}{a^{2}}\) - \(\frac{o^{2}}{a^{2}}\).

                                      = \(\frac{h^{2} - o^{2}}{a^{2}}\)

                                      = \(\frac{a^{2}}{a^{2}}\); [Since, in the right-angled ∆XYZ, o\(^{2}\) + a\(^{2}\) = h\(^{2}\) ⟹ h\(^{2}\) - o\(^{2}\) = a\(^{2}\) (by Pythagoras’ Theorem)]

                                      = 1.

Therefore, sec\(^{2}\) θ - tan\(^{2}\) θ = 1.

Consequently, 1 + tan\(^{2}\) θ = sec\(^{2}\) θ and sec\(^{2}\) θ - 1 = tan\(^{2}\) θ.

 

 

III. csc\(^{2}\) θ - cot\(^{2}\) θ = 1

We have, csc θ = \(\frac{h}{o}\) and cot θ = \(\frac{a}{o}\).

Therefore, csc\(^{2}\) θ - cot\(^{2}\) θ = \(\frac{h^{2}}{o^{2}}\) - \(\frac{a^{2}}{o^{2}}\).

                                       = \(\frac{h^{2} - a^{2}}{o^{2}}\)

                                       = \(\frac{o^{2}}{o^{2}}\); [Since, in the right-angled ∆XYZ, o\(^{2}\) + a\(^{2}\) = h\(^{2}\) ⟹ h\(^{2}\) - a\(^{2}\) = o\(^{2}\) (by Pythagoras’ Theorem)]

                                       = 1.

Therefore, csc\(^{2}\) θ - cot\(^{2}\) θ = 1.

Consequently, 1 + cot\(^{2}\) θ = csc\(^{2}\) θ and csc\(^{2}\) θ - 1 = cot\(^{2}\) θ.

These three trigonometric identities are also called Pythagorean identities.

Pythagorean Identities

Note: The equalities csc θ = \(\frac{1}{sin θ}\), sec θ = \(\frac{1}{cos θ}\), cot θ = \(\frac{1}{tan θ}\), tan θ = \(\frac{sin θ}{cos θ}\) and cot θ = \(\frac{cos θ}{sin θ}\) are holds for all values of θ. Therefore, these equalities are trigonometric identities.

Thus, we have the following trigonometric identities.


Table of Trigonometric Identities

1. csc θ = \(\frac{1}{sin θ}\),

2. sec θ = \(\frac{1}{cos θ}\),

3. cot θ = \(\frac{1}{tan θ}\),

4. tan θ = \(\frac{sin θ}{cos θ}\),

5. cot θ = \(\frac{cos θ}{sin θ}\)

6. sin\(^{2}\) θ + cos\(^{2}\) θ = 1; 1 - sin\(^{2}\) θ = cos\(^{2}\) θ, 1 - cos\(^{2}\) θ = sin\(^{2}\) θ.

7. sec\(^{2}\) θ - tan\(^{2}\) θ = 1; 1 + tan\(^{2}\) θ = sec\(^{2}\) θ; sec\(^{2}\) θ - 1 = tan\(^{2}\) θ.

8. csc\(^{2}\) θ - cot\(^{2}\) θ = 1

9. 1 + cot\(^{2}\) θ = csc\(^{2}\) θ,

10. csc\(^{2}\) θ - 1 = cot\(^{2}\) θ.

Table of Trigonometric Identities

Solved Examples on Trigonometric Identities:

1. Prove that sin\(^{4}\) θ + cos\(^{4}\)  θ + 2 sin\(^{2}\) θ ∙ cos\(^{2}\)  θ = 1

Solution:

LHS = sin\(^{4}\) θ + cos\(^{4}\)  θ + 2 sin\(^{2}\) θ ∙ cos\(^{2}\)  θ

       = (sin\(^{2}\) θ)\(^{2}\) + (cos\(^{2}\)  θ)\(^{2}\) + 2 sin\(^{2}\) θ ∙ cos\(^{2}\)  θ

       = (sin\(^{2}\) θ + cos\(^{2}\) θ)\(^{2}\)

       = 1\(^{2}\); [Since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

       = 1 = RHS. (Proved).


How to Solve Trigonometric Identities Proving Problems?

2. Show that sec θ - cos θ = sin θ tan θ

Solution:

LHS =  sec θ - cos θ

      = \(\frac{1}{cos θ}\) - cos θ

      = \(\frac{1 - cos^{2} θ}{cos θ}\)

      = \(\frac{sin^{2} θ}{cos θ}\);  [Since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

      = sin θ ∙ \(\frac{sin θ}{cos θ}\)

      = sin θ ∙ tan θ = RHS. (Proved)


3. Prove that 1 - \(\frac{cos^{2} A}{1 + sin A}\) = sin A

Solution:

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

       = 1 - \(\frac{1 - sin^{2} A}{1 + sin A}\); [cos\(^{2}\) θ = 1 - sin\(^{2}\) θ]

       = 1 - \(\frac{(1 + sin A)(1 – sin A)}{1 + sin A}\)

       = 1 – (1 – sin A)

       = 1 – 1 + sin A

       = sin A = RHS. (Proved).


Verifying Trigonometric Identities & Equations

4. Prove that \(\frac{sin A}{1 + cos A}\) + \(\frac{sin A}{1 - cos A}\) = 2 csc A.

Solution:

LHS = \(\frac{sin A}{1 + cos A}\) + \(\frac{sin A}{1 - cos A}\)

       = \(\frac{sin A(1 – cos A) + sin A(1 + cos A)}{(1 + cos A)(1 – cos A)}\)

       = \(\frac{sin A – sin A ∙ cos A + sin A + sin A ∙ cos A)}{1 - cos^{2} A}\)

       = \(\frac{2 sin A}{sin^{2} A}\); [Since, 1 - cos\(^{2}\) A = sin\(^{2}\) A]

       = \(\frac{2}{sin A}\)

       = 2 ∙ \(\frac{1}{sin A}\); [Since, csc A = \(\frac{1}{sin A}\)]

       = RHS. (Proved).


Proving Trigonometric Identities Practice Problems Online

5. Prove that \(\frac{sin A}{1 + cos A}\) = \(\frac{1 – cos A}{sin A}\).

Solution:

LHS = \(\frac{sin A}{1 + cos A}\)

       = \(\frac{sin A}{1 + cos A}\) ∙ \(\frac{1 - cos A}{1 - cos A}\)

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

       = \(\frac{sin A(1 - cos A)}{sin^{2} A}\); [Since 1 - cos\(^{2}\) θ = sin\(^{2}\) θ]

       = \(\frac{1 - cos A}{sin A}\) = RHS. (Proved).


Trigonometry Problems and Questions with Solutions

6. Prove that \(\sqrt{\frac{1 + sin θ}{1 - sin θ}}\) = sec θ +tan θ.

Solution:

LHS = \(\sqrt{\frac{1 + sin θ}{1 - sin θ}}\)

       = \(\sqrt{\frac{(1 + sin θ) (1 + sin θ)}{(1 - sin θ)(1 + sin θ)}}\)

       = \(\sqrt{\frac{(1 + sin θ)^{2}}{1 – sin^{2} θ}}\)

       = \(\sqrt{\frac{(1 + sin θ)^{2}}{cos^{2} θ}}\); [Since 1 - sin\(^{2}\) θ = cos\(^{2}\) θ]

       = \(\frac{1 + sin θ}{cos θ}\)

       = \(\frac{1}{cos θ}\) + \(\frac{sin θ}{cos θ}\)

       = secθ + tan θ = RHS. (Proved).


Trigonometric identities problems for class 10

7. Prove that tan\(^{2}\) A – tan\(^{2}\) B = \(\frac{sin^{2} A - sin^{2} B}{cos^{2} A ∙ cos^{2} B}\) = sec\(^{2}\) A - sec\(^{2}\) B

Solution:

LHS = tan\(^{2}\) A – tan\(^{2}\) B

       = \(\frac{sin^{2} A}{cos^{2} A}\) - \(\frac{ sin^{2} B}{cos^{2} B}\)

       = \(\frac{sin^{2} A ∙ cos^{2} B - sin^{2} B ∙ cos^{2} A}{cos^{2} A ∙ cos^{2} B}\)

       = \(\frac{sin^{2} A (1 - sin^{2} B) - sin^{2} B(1 - sin^{2} A)}{cos^{2} A ∙ cos^{2} B}\); [Since, cos2 B = 1 - sin2 B and cos2 A = 1 - sin2 A]

       = \(\frac{sin^{2} A  - sin^{2} A  sin^{2} B - sin^{2} B + sin^{2} B sin^{2} A}{cos^{2} A ∙ cos^{2} B}\);

        = \(\frac{sin^{2} A - sin^{2} B}{cos^{2} A ∙ cos^{2} B}\) = MHS

        = \(\frac{(1 - cos^{2} A) - (1 - cos^{2} B)}{cos^{2} A ∙ cos^{2} B}\); [Since, sin2 A = 1 - cos2 A and sin2 B = 1 - cos2 B]

        = \(\frac{cos^{2} B - cos^{2} A}{cos^{2} A ∙ cos^{2} B}\)

        = \(\frac{cos^{2} B}{cos^{2} A ∙ cos^{2} B}\) - \(\frac{cos^{2} A}{cos^{2} A ∙ cos^{2} B}\)

        = \(\frac{1}{cos^{2} A}\) - \(\frac{1}{cos^{2} B}\)

        = sec\(^{2}\) A – sec\(^{2}\) B = RHS. (Proved)

Again, 

LHS = tan\(^{2}\) A – tan\(^{2}\) B

       = (sec\(^{2}\) A – 1) – (sec\(^{2}\) B – 1); [Since, tan\(^{2}\) θ = sec\(^{2}\) θ – 1 ]

       = sec\(^{2}\) A – 1 – (sec\(^{2}\) B +1

       = sec\(^{2}\) A – sec\(^{2}\) B = RHS. (Proved).


Solving Trigonometric Equations using Trigonometric Identities

8. Prove that tan\(^{2}\) θ - sin\(^{2}\) θ = tan\(^{2}\) θ ∙ sin\(^{2}\) θ

Solution:

LHS = tan\(^{2}\) θ - sin\(^{2}\) θ

       = tan\(^{2}\) θ - \(\frac{sin^{2} θ}{cos^{2} θ}\) ∙ cos\(^{2}\)

       = tan\(^{2}\) θ - tan\(^{2}\) θ ∙ cos\(^{2}\)

       = tan\(^{2}\) θ(1 - cos\(^{2}\))

       = tan\(^{2}\) θ ∙ sin\(^{2}\) θ = RHS. (Proved).


List of trigonometric identities Problems

9. Prove that 1 + \(\frac{cot^{2} θ}{1 + csc θ}\) = csc θ

Solution:

LHS = 1 + \(\frac{cot^{2} θ}{1 + csc θ}\)

       = 1 + \(\frac{csc^{2} θ - 1}{1 + csc θ}\)

       = 1 + \(\frac{(csc θ + 1)(csc θ – 1)}{1 + csc θ}\)

       = 1 + (csc θ – 1)

       = 1 + csc θ – 1

       = csc θ = RHS. (Proved).


Solve the problem using trigonometric identities

10. Prove that \(\frac{sec θ - )}{sec θ + 1}\) = (cot θ – csc θ)\(^{2}\).

Solution:

LHS = \(\frac{sec θ - 1}{sec θ + 1}\)

       = \(\frac{sec θ - 1}{sec θ + 1}\) ∙ \(\frac{sec θ - 1}{sec θ - 1}\)

       = \(\frac{(sec θ - 1)^{2}}{sec^{2} θ - 1}\)

       = \(\frac{(sec θ - 1)^{2}}{tan^{2} θ}\); [Since sec\(^{2}\) θ – 1 = tan\(^{2}\) θ]

       = \((\frac{sec θ - 1}{tan θ})^{2}\)

       = \((\frac{\frac{1}{cos θ} - 1}{\frac{sin θ }{cos θ} })^{2}\)

       = \((\frac{1 – cos θ}{cos θ} ∙ \frac{cos θ}{sin θ})^{2}\)

       = \((\frac{1 – cos θ}{sin θ})^{2}\)

       = \((\frac{1}{sin θ} - \frac{cos θ}{sin θ})^{2}\)

       = (csc θ – cot θ)\(^{2}\)

       = (cot θ – csc θ)\(^{2}\) = RHS. (Proved).


Proving Trigonometric Identities

11. Prove that \(\frac{sin A}{cot A + csc A}\) = 2 + \(\frac{sin A}{cot A - csc A }\)

Solution:

LHS = \(\frac{sin A}{cot A + csc A}\)

       = \(\frac{sin A(cot A - csc A)}{(cot A + csc A)(cot A - csc A)}\)

       = \(\frac{sin A ∙ cot A – sin A ∙ csc A}{cot^{2} A – csc^{2} A}\)

       = \(\frac{sin A ∙ \frac{cos A}{sin A} – 1}{-1}\); [since, sin θ csc θ = 1, csc\(^{2}\) θ – cot\(^{2}\)  θ = 1]

       = \(\frac{cos A - 1}{-1}\)

       = 1 – cos A

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

        = 2 + \(\frac{sin A}{ cot A - csc A}\) ∙ \(\frac{(cot A + csc A)}{(cot A + csc A)}\)

        = 2 + \(\frac{sin A(cot A + csc A)}{cot^{2} A – csc^{2} A}\)

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

        = 2 + \(\frac{cos A + 1}{-1}\)

        = 2 – (cos A + 1)

        = 2 – cos A – 1

        = 1 – cos A

Therefore, LHS = RHS. (Proved).


Alternative Method

The identity is established if we can prove that

\(\frac{sin A}{cot A + csc A}\) - \(\frac{sin A}{cot A - csc A }\) = 2

Now, here  

LHS = \(\frac{sin A}{cot A + csc A}\) - \(\frac{sin A}{cot A - csc A }\)

       = \(\frac{sin A(cot A - csc A) – sin A(cot A + csc A)}{( cot A + csc A)(cot A - csc A)}\)

       = \(\frac{sin A ∙ cot A – sin A ∙ csc A – sin A ∙ cot A – sin A ∙ csc A}{(cot A + csc A)(cot A - csc A)}\)

       = \(\frac{–2 sin A ∙ csc A}{cot^{2} A – csc^{2} A}\)

       = \(\frac{–2}{-1 }\); [Since sin θ csc θ = 1, csc\(^{2}\) θ – cot\(^{2}\)  θ = 1]

       = 2 = RHS. (Proved)


Problems on Evaluation using Trigonometric Identities:

12. If sin θ + csc θ = 2, find the value of sin\(^{10}\) θ + csc\(^{11}\) θ

Solution:

Given that, sin θ + csc θ = 2 ……………. (i)

⟹ sin θ + \(\frac{1}{ sin θ}\) = 2

⟹ \(\frac{ sin^{2} θ + 1}{sin θ }\) = 2

⟹ sin\(^{2}\) θ + 1= 2 sin θ

⟹ sin\(^{2}\) θ - 2 sin θ + 1 = 0

⟹ (sin θ - 1)\(^{2}\) = 0

⟹ sin θ - 1 = 0

⟹ sin θ = 1

⟹ csc θ = \(\frac{1}{ sin θ}\) = \(\frac{1}{ 1}\) = 1

Therefore, csc θ = 1.

Now, sin\(^{10}\) θ + csc\(^{11}\) θ

= 1\(^{10}\) + 1\(^{11}\)

= 1 + 1

= 2.

 

13. If sec θ – tan θ = \(\frac{1}{3}\), find the value of sec A and tan A.

Solution:

Given, sec θ – tan θ = \(\frac{1}{3}\) ………… (i)

We know that the Pythagorean identity, sec\(^{2}\) θ - tan\(^{2}\) θ = 1.

                   ⟹ (sec θ + tan θ)(sec θ - tan θ) = 1

                   ⟹ (sec θ + tan θ) ∙ \(\frac{1}{3}\) = 1, [Given sec θ – tan θ = \(\frac{1}{3}\)]

                   ⟹ sec θ + tan θ = 3 ……….. (ii)

Now adding (i) and (ii) we get

                       2 sec θ = \(\frac{1}{3}\) + 3

                   ⟹ 2 sec θ = \(\frac{10}{3}\)

                   ⟹ sec θ = \(\frac{10}{3}\) ∙ \(\frac{1}{2}\)

                   ⟹ sec θ = \(\frac{5}{3}\)

Therefore, sec θ = \(\frac{5}{3}\)

Putting the value of sec θ = \(\frac{5}{3}\) in (ii), we get

\(\frac{5}{3}\) + tan θ = 3

⟹ tan θ = 3 - \(\frac{5}{3}\)

⟹ tan θ = \(\frac{4}{3}\)

Therefore, tan θ = \(\frac{4}{3}\).


14. If tan A + sec A = \(\frac{2}{\sqrt{3}}\), find the value of sin A

Solution:

Given that, tan A + sec A = \(\frac{2}{\sqrt{3}}\) ……………….. (i)

We know the Pythagorean trigonometric identity,

       sec\(^{2}\) A - tan\(^{2}\) A = 1.

⟹ (sec A + tan A)(sec A – tan A) = 1

⟹ \(\frac{2}{\sqrt{3}}\)(sec A – tan A) = 1; [given , tan A + sec A = \(\frac{2}{\sqrt{3}}\)]

⟹ sec A – tan A = \(\frac{\sqrt{3}}{2}\) ……………….. (ii)


Solving these two equations we get,

     2 sec A = \(\frac{2}{\sqrt{3}}\) + \(\frac{\sqrt{3}}{2}\)

 ⟹ 2 sec A = \(\frac{4 + 3}{2\sqrt{3}}\)

 ⟹ 2 sec A = \(\frac{7}{2\sqrt{3}}\)

⟹ sec A = \(\frac{7}{4\sqrt{3}}\) ……………….. (iii)


Putting the value of sec A = \(\frac{7}{4\sqrt{3}}\) in (i) we get,

     tan A + \(\frac{7}{4\sqrt{3}}\) = \(\frac{2}{\sqrt{3}}\)

⟹ tan A = \(\frac{2}{\sqrt{3}}\) - \(\frac{7}{4\sqrt{3}}\)

⟹ tan A = \(\frac{8 - 7}{4\sqrt{3}}\)

⟹ tan A = \(\frac{1}{4\sqrt{3}}\)……………….. (iv)


Now dividing (iv) by (iii) we get,

\(\frac{tan A}{sec A}\) = \(\frac{1}{7}\)

⟹ sin A = \(\frac{1}{7}\)

Therefore, sin A = \(\frac{1}{7}\).


Problems on establishing conditional results

15. If cos A + sin A = \(\sqrt{2}\) cos A, prove that cos A - sin A = \(\sqrt{2}\) sin A.

Solution:

Let cos A - sin A = k. Now,

(cos A + sin A)\(^{2}\) + (cos A - sin A)\(^{2}\) = cos\(^{2}\) A + sin\(^{2}\) A + 2 cos A ∙ sin A + cos\(^{2}\) A + sin\(^{2}\) A - 2 cos A ∙ sin A

                                            = 1 + 2 cos A sin A + 1 - 2 cos A sin A

                                            = 2

Therefore, (\(\sqrt{2}\) cos A)\(^{2}\) + k\(^{2}\) = 2

⟹ k\(^{2}\) = 2 - (\(\sqrt{2}\) cos A)\(^{2}\) 

⟹ k\(^{2}\) = 2 - 2cos\(^{2}\) A

⟹ k\(^{2}\) = 2(1 - cos\(^{2}\) A)

⟹ k\(^{2}\) = 2sin\(^{2}\) A; [Since, sin\(^{2}\) θ + cos\(^{2}\) θ = 1]

⟹ k = \(\sqrt{2}\) sin A

⟹ cos A - sin A = \(\sqrt{2}\) sin A. (Proved).





10th Grade Math

From Trigonometric Identities to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Worksheet on Triangle | Homework on Triangle | Different types|Answers

    Jun 21, 24 02:19 AM

    Find the Number of Triangles
    In the worksheet on triangle we will solve 12 different types of questions. 1. Take three non - collinear points L, M, N. Join LM, MN and NL. What figure do you get? Name: (a)The side opposite to ∠L…

    Read More

  2. Worksheet on Circle |Homework on Circle |Questions on Circle |Problems

    Jun 21, 24 01:59 AM

    Circle
    In worksheet on circle we will solve 10 different types of question in circle. 1. The following figure shows a circle with centre O and some line segments drawn in it. Classify the line segments as ra…

    Read More

  3. Circle Math | Parts of a Circle | Terms Related to the Circle | Symbol

    Jun 21, 24 01:30 AM

    Circle using a Compass
    In circle math the terms related to the circle are discussed here. A circle is such a closed curve whose every point is equidistant from a fixed point called its centre. The symbol of circle is O. We…

    Read More

  4. Circle | Interior and Exterior of a Circle | Radius|Problems on Circle

    Jun 21, 24 01:00 AM

    Semi-circular Region
    A circle is the set of all those point in a plane whose distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is known

    Read More

  5. Quadrilateral Worksheet |Different Types of Questions in Quadrilateral

    Jun 19, 24 09:49 AM

    In math practice test on quadrilateral worksheet we will practice different types of questions in quadrilateral. Students can practice the questions of quadrilateral worksheet before the examinations

    Read More