# alpha

Secø + tanø=x
Then find the value of secø

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1. 1/cosø + sinø/cosø = x
(1+sinø)/cosø = x
(1+sinø)^2/(1-sin^2ø) = x^2
1+2sinø+sin^2ø = x^2 - x^2sin^2ø
(1+x^2)sin^2ø + 2sinø +(1-x^2) = 0

Now you can use the quadratic formula to express sinø in terms of x. Then, of course,

secø = 1/cosø = 1/sqrt(1-sin^2ø)

You sure you typed what you intended?

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2. secØ = x - tanØ for a trivial solution, but ....

If you want secØ in terms of x, then you have a mess

Secø + tanø=x
1/cosØ + sinØ/cosØ = x
(1+sinØ)/cosØ = x
1 + sinØ = xcosØ
square both sides
1 + 2sinØ + sin^2 Ø = x^2 cos^2 Ø
1 + 2sinØ + sin^2 Ø = x^2(1-sin^2 Ø)
1 + 2sinØ + sin^2 Ø = x^2 - x^2 sin^2 Ø

(x^2 + 1) sin^2 Ø + 2sinØ + 1 - x^2 = 0
a quadratic in sinØ , where
a = x^2 + 1
b = 2
c = 1 - x^2

so we could find sinØ
= (-2 ± √(4 - 4(x^2 - 1)(1-x^2) )/(2(x^2 + 1))
= (-2 ± √(4 + 4(x^2 - 1)(x^2 + 1) )/(2(x^2 + 1))
= (-2 ± 2√(4 + 4x^4 - 4)/(2(x^2 + 1))
= (-1 ± √x^4) )/(x^2 + 1)
= (-1 + x^2)/(x^2 + 1) or (-1 - x^2)/(x^2 + 1)
= (x^2 - 1)/(x^2 + 1) or -1
but if sinØ = -1, then the original secØ would be undefined, so

sinØ = (x^2-1)/(x^2 + 1)

arggghhhh, but we wanted secØ

cos^2 Ø = 1 - sin^2 Ø
= 1 - (x^2 - 1)^2 / (x^2 + 1)^2
= ( x^4 + 2x^2 + 1 - x^4 + 2x^2 - 1)/(x^2 + 1)^2
= 4x^2/(x^2 + 1)^2
cosØ = 2x/(x^2 + 1)

secØ = (x^2 + 1)/(2x)

ok, ok, it is now easy to see that we are dealing with a right-angled triangle with legs 2x and x^2 -1 and hypotenuse x^2 + 1
and I can easily verify my answer:

LS = secØ + tanØ
= (x^2 + 1)/(2x) + (x^2 - 1)/(2x)
= 2x^2/(2x)
= x
= RS

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3. We know sec²ø-tan²ø=1
Therefore (secø+tanø)(secø-tanø)=1
From this we get secø-tanø=1/x,secø+tanø=x
Subtract both equation
2secø=x+1/x
Secø=1/2(x+1/x)

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