A level trignometry question?

• prg

  22 Dec 08, 14:37

  The angle A is such that secA+tanA=2

  show that secA - tanA=1/2 and hence find exact value of cos A

• eagle

  22 Dec 08, 14:40

  sec² A = 1 + tan² A
  sec² A - tan² A = 1
  (sec A + tan A)(sec A - tan A) = 1
  (2) (sec A - tan A) = 1
  secA - tanA=1/2 (shown)

  secA - tanA + secA+tanA = 2.5
  sec A = 1.25
  cos A = 0.8 = 36.9 degrees

• prg

  22 Dec 08, 21:40

  ok thanks,

  Some more questions

  (a) Prove identity cot(1/2)A - tan (1/2)A= 2cotA

  **i have attempted it **

  1/ tan(1/2)A - tan(1/2)A=(1 – tan²(A/2)) / (tanA/2)

  If tan2A=(2tanA) / 1-tan²A   ----(1)

   Let A = A/2

  tan A =  ( 2tan(A/2) ) / (1 – tan²(A/2)) 

  1 – tan²(A/2) =  ( 2tan(A/2) ) / (tan A )    ----(2)

   from (2)

   (1 – tan²(A/2)) / (tanA/2) = ( 2tan(A/2) ) / (tanA/2) (tan A ) = 2 / tanA

   1 / tanA =cot A

   Therefore, cot(1/2)A - tan (1/2)A= 2cotA

  **is this correct way, is there any easier way??**

   

  (b)By choosing suitable numerical value for A show that tan 15 is a root of quadratic equation t^2 + (2√3)t  - 1 =0

   

  **how to do this??**

   

• wee_ws

  22 Dec 08, 22:20

  Hi,

    For the proof, a faster way is to express cot and tan in terms of sin and cos, i.e.

    (cos A/2) / (sin A/2) - (sin A/2) / (cos A/2)

    = [ cos^2 A/2 - sin^2 A/2 ] / [ sin A/2 cos A/2 ]

    = cos A / [ 1/2 sin A ]

    = 2 cot A (shown)

    Let A = 15 deg, since we know that tan 30 deg = 1/sqrt(3). Then we obtain

    cot 15 - tan 15 = 2 cot 30

    1 / (tan 15) - tan 15 = 2 sqrt(3)

    Multiplying both sides by tan 15, we obtain

    1 - tan^2 15 = 2 sqrt(3) (tan 15)

    Rearranging, we have tan^2 15 + (2 sqrt(3)) (tan 15) - 1 = 0, of which we see that t = tan 15 is a root to the said quadratic equation.

    Thanks!

  Cheers,
  Wen Shih