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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.
 

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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

In free Live math tutoring online the students can avail opportunity to learn the concepts online and enhance their Knowledge without spending much time in going through the books.Here we have solved some problems
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
x2 + y2 + 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 x2 + y2 + 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 x4 - 17x2 + 52 = 0

Solution: -

Put x2 = u then the given equation becomes u2 - 17u + 52 = 0
Splitting the middle term,
u2 - 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
x2 =13 Then x = ± √13
when u = 4
x2 = 4 Then x = ± 2
Hence the roots of the given equation are ± √13, ± 2