Live math tutoring online is a method of getting math tutoring help from our tutors. This will help you to learn and master math topics. Our tutors are available 24 x 7 to help you in a simple, clear and understanding method. We can assure that you can see dramatic improvements in your grades. We also offer help on

assignments and home work problems in math. We give interactive lesions in Math, worksheets and homework help and is designed in such a way that you can master each an every topic. So what are you waiting for? Avail help as early as possible and master the different topics in math.

## Free Live Math Tutoring

Following are few topics on which live tutors are available.

assignments and home work problems in math. We give interactive lesions in Math, worksheets and homework help and is designed in such a way that you can master each an every topic. So what are you waiting for? Avail help as early as possible and master the different topics in math.

## Free Live Math Tutoring

Following are few topics on which live tutors are available.

- Pre Algebra
- Algebra
- Number Sense
- Calculus
- Trigonometry
- Geometry
- Probability
- Statistics.

## Free Live Math Tutoring Online

Following are the samples of solved problems by our live tutors.

**Example 1: -**

If A + B + C = 180, prove that sin 2A + sin 2B - sin 2C = 4 cos A cos B sin C

**Solution: -**

sin 2A + sin 2B - sin 2C = (sin 2A + sin 2B) - sin 2C

= 2 sin (A + B) cos (A - B) - 2 sin C cos C

= 2 sin (180 - C) cos (A - B) - 2 sin C cos C

= 2 sin C cos (A - B) - 2 sin C cos C

Taking 2 sin C common, we get

sin 2A + sin 2B - sin 2C = 2 sin C [cos (A - B) - cos C]

But A + B + C = 180. So C = 180 - (A + B)

Therefore we get

in 2A + sin 2B - sin 2C = 2 sin C [cos (A - B) - cos (180 - (A + B))]

= 2 sin C [cos (A - B) + cos (A + B)]

= 2 sin C 2 cos A cos B

= 4 cos A cos B sin C

Therefore

**sin 2A + sin 2B - sin 2C = 4 cos A cos B sin C**

**Example 2 : -**

Show that the points (-3, 11), (5, 9), (8, 0) and (6, 8) are the vertices of a cyclic quadrilateral.

**Solution: -**

Let the circle passing through the first three points be

x

^{2 }+ y

^{2}+ 2gx + 2fy + c = 0...(1)

Substituting the first three point in the above equation, we get the following three equations.

9 + 121 - 6g + 22f + c = 0

25 + 81 + 10g + 18f + c = 0

64 + 16g + c = 0

That is

-6g + 22f + c = -130 ...(2)

10g + 18f + c = -106 ...(3)

16g + c = -64 ...(4)

Solving (2), (3), and (4), we get

g = 1, f = -2 and c = -80.

So the circle passing through the three point is x

^{2}+ y

^{2}+ 2x - 4y - 80 = 0.

Now we substitute the last point.

Putting x = 6 and y = 8 in this equation, we get

36 + 64 + 12 - 32 - 80 = 0

That is (6, 8) lies on the circle passing through the first three points.

**Therefore the given points are the vertices of a cyclic quadrilateral.**

**Example 3: -**

Solve x

^{4}- 17x

^{2}+ 52 = 0

**Solution: -**

Put x

^{2}= u then the given equation becomes u

^{2}- 17u + 52 = 0

Splitting the middle term,

u

^{2}- 13u -4u + 52 = 0

taking u common from first two terms and -4 common from last two terms we get

u(u -13) - 4(u -13) =0

(u -13)(u - 4) = 0

So u =13 or u = 4

when u = 13

x

^{2}=13 Then x = ± √13

when u = 4

x

^{2}= 4 Then x = ± 2

**Hence the roots of the given equation are ± √13, ± 2**