Question

Prove that 4sin50 degree minus root 3tan50 degree = 1

Answer 1

Given: The term  4sin50 degree minus root 3tan50 degree = 1

To find: Prove LHS = RHS.

Solution:

• Now we have given: 4sin50° - √3tan50° = 1

• Consider LHS, we have:

             4sin50° - √3tan50°

• We can rewrite it as:

             4sin50° - √3sin50°/cos50°

             ( 4sin50.cos50° - √3sin50°) / cos50°

• Now we know the formula:

             sin2A = 2sinAcosA

• So applying this, we get:

             (2sin100° - √3sin50°) / cos50°

             100 can be written as 180 - 80, so:

             2{ sin(180°-80°) - √3/2 sin50°} / cos50°

             2{ sin80° - cos30° x sin50°}cos50°

             2{ sin80° -1/2(2sin50° x cos30°) } cos50°

• Now we know the formula:

             2sinA x cosB = sin( A+B) + sin(A - B)

             So applying this, we get:

             2{ sin80° -1/2( sin80° + sin20°)}/cos50°

             2{ 1/2sin80° - 1/2sin20°}/cos50°

             ( sin80° - sin20°)/cos50°

• Again using the formula:

             sinA - sinB = 2cos( A + B)/2.sin(A - B)/2

             {2cos( 80+20)/2 x sin(80-20)/2 }/cos50°

             2cos50° x sin30°/cos50°

             2sin30°

             2 x 1/2

             1

             RHS.

• Hence proved.

Answer:

          So we have proved 4sin50° - √3tan50° = 1