A level trignometry question?

The angle A is such that secA+tanA=2

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

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

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??**

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