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hey ppl, sorry but I have solved nearly all the trig questions in the mixed exercise, ranging from easy to difficult. However, ther have been two questions that I have been finding immensely difficult to solve, which is beginning to bug me since I am sure I have been able to do them before, but anyway here they are:

17) Given that sec x + tan x = -3 , u se the identity 1 + tan^2x = sec^2x to find the value of sec x - tan x

this one is the worst:

Given that p = sec x - tan x and q = sec x + tan x, show that p = 1/q??

please help, thanks guys!

17) Given that sec x + tan x = -3 , u se the identity 1 + tan^2x = sec^2x to find the value of sec x - tan x

this one is the worst:

Given that p = sec x - tan x and q = sec x + tan x, show that p = 1/q??

please help, thanks guys!

I will complete the second question first as it helps with the first one:

If p = 1/q then pq = 1

p = (secx - tanx)

q = (secx + tanx)

pq = (secx - tanx)(secx + tanx)

= sec^2(x) + secxtanx - secxtanx - tan^2(x)

= sec^2(x) - tan^2(x)

sec^2(x) = 1 + tan^2(x), substituting this in:

= (1 + tan^2(x)) - tan^2(x)

= 1

Q.E.D. Thus p = 1/q

(secx + tanx) = -3

From above we have:

1/(secx + tanx) = (secx - tanx)

(q = 1/p)

So:

-1/3 = (secx - tanx)

If p = 1/q then pq = 1

p = (secx - tanx)

q = (secx + tanx)

pq = (secx - tanx)(secx + tanx)

= sec^2(x) + secxtanx - secxtanx - tan^2(x)

= sec^2(x) - tan^2(x)

sec^2(x) = 1 + tan^2(x), substituting this in:

= (1 + tan^2(x)) - tan^2(x)

= 1

Q.E.D. Thus p = 1/q

(secx + tanx) = -3

From above we have:

1/(secx + tanx) = (secx - tanx)

(q = 1/p)

So:

-1/3 = (secx - tanx)

SunGod87

I already do Astrophysics at University, but maybe when I finish I'll take up a Maths degree!

Wow! that sounds interesting, i'm more interested in chem tbh, but maths is great! Btw, do you use a lot of trig in what u r studying?

O yeh, I managed to work out the answer to number 17, this way?

From the identity sec^2x = 1 + tan^2x

It follows that (-3-tanx)^2 = sec^2 x and so (-3-tanx)^ 2 - tan^2 x = 1

this gives : 9 + 6tanx = 1

6tanx = -8

tan x = -4/3 From this I can solve sec x - tan x:

As I said sec x = -3-tan x

sec x - tan x = -3-(-4/3)-(-4/3) which equates to -1/3 yerrr! lol

Do you think my method is too complicated or do u reckon it it fine?

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can someone please explain what principle domain is and why the answer is a not c?Maths

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