Question which the teacher cant help with

[bob4000]

a)state the value of sec^2x-tan^2x
b) the angle A is such that secA + tanA = 2. show that secA-tanA=1/2, and hence find the exact value of cos A.

part a was a doddle so to speak, however, part b was on the contrary; it took our class half an hour, but to avail

if someone could shed light on this problem, i would be much obliged

thank you in advance

 

[HallsofIvy]

Not all that difficult, surely.
[tex] sec A= \frac{1}{cos A}[/tex] and [tex]tan A\frac{sin A}{cos A}[/tex]
so [tex]sec A+ tan A= \frac{1+ sin A}{cos A}= 2[/tex]
Looks to me like an obvious thing to try is to multiply numerator and denominator of that equation by 1- sin A:
[tex]\frac{(1+ sin A)(1- sin A)}{(cos A)(1- sin A)}= 2[/tex]
[tex]\frac{1- sin^2 A}{(cos A)(1- sin A)}= \frac{cos^2 A}{(cos A)(1- sin A)}= \frac{cos A}{1- sin A}= 2[/tex]
Then
[tex] \frac{1- sin A}{cos A}= \frac{1}{2}[/tex]
[tex] \frac{1}{cos A}- \frac{sin A}{cos A}= sec A- tan A= \frac{1}{2}[/tex].
Add the two equations to eliminate tan A and then invert.

 

[bob4000]

thanks. i think we were all just looking at it form the wron perspective. much obliged