Question

If sec A =17/8 , verify that 3-4sin²A/4cos²A-3=3-tan²A/1-3tan²A

Answer 1

3-4sin²A / 4cos²A-3 = 3-tan²A / 1-3tan²A  proved

Step-by-step explanation:

Given: sec A =17/8

To prove: prove that 3-4sin²A / 4cos²A-3 = 3-tan²A / 1-3tan²A

Solution:  

 SecA = 17/8 = Hypotenuse / Adjacent  

So Opposite = √(17² - 8² = √(289 - 64) = √225 = 15

Now Sin A = opposite / hypotenuse = 15/17

Cos A = Adjacent / Hypotenuse  = 8/17

Tan A = Opposite / Adjacent = 15/8

So LHS = 3-4sin²A / 4cos²A-3

      = [3 - 4(15/17)²] / [4(8/17)² -3]

      = 3 -4(225)/289] / 4(64)/289 - 3

      = [3(289) - 4(225)] / 4(64) - 3(289)

      = 867 - 900 / 256 - 867

      = -33 / -611

      = 33/611

RHS = 3-tan²A / 1-3tan²A

    = 3 -(15/8)² / 1-3(15/8)²

    = 3 - 225/64 / 1 -3(225/64)

    = 3(64) - 225 / 64 - 3(225)

    = 192 -225 / 64 - 675

    = -33 /-611

    = 33/611

  LHS = RHS

Hence proved.