Could someone please help? I don't know how to solve this - trigonometry?

Question

If secQ ( theta ) = x + 1/(4x), prove that secQ+ tanQ = 2x, or 1/(2x)

This isn't a H.W. question. I have a test and it's kind of urgent

Answer 1

sec2Q - Tan2Q= 1 (sec2Q = sec square Theta)
or Tan2Q= sec2 Q - 1
or Tan2 Q = ((4x2+1)/4x)2 - 1
or Tan2Q = (16x4 + 8x2+1- 16x2)/16x2
or tanQ = +or -(4x2-1)/4x
or tanQ = + or - (x - 1/4x)
Now,
secQ + tanQ = ( x+ 1/4x) + or - (x- 1/4x)
= 2x or 1/2x

therefore secQ+ TanQ = 2x or 1/2x

Answer 2

secQ ( theta ) = x + 1/(4x), prove that secQ+ tanQ = 2x, or 1/(2x)

Note : As in this note pad format we cant type Theta. I am using A instead of Theta.

Given: --- sec A = x + 1/ (4 x ) .
To prove that secA+ tanA = 2x, or 1/(2x)

Soln:

Given : ---
...........................................
Sec A = x + 1(4x) ....................................... (1)
............................................
Squaring both the sides we get
sec^2 A = [ x + 1(4x) ]^2
=> sec^2 A = x^2 + 1/(16x^2) + (2x)/(4x)
=> sec^2 A = x^2 + 1/(16x^2) + ½
Subtract 1 from both the sides of the identity.
=> sec^2 A - 1 = x^2 + 1/(16x^2) + ½ - 1
.......................................................................
[ Note that sec^2 A - 1 = tan^2 A
......................................................................................
=> tan^2 A = x^2 A + 1/(16x^2) - ½
=> tan^2 A = ( x - 1(4x) ]^2
Now take sqrt of both the sides to get ---
.......................................................................
=> tan A = (+/-) [ x - 1(4x) ] --------------- (2)
.....................................................................

From (1) and (2) we get-

sec A + tan A = [ x + 1(4x)] +/- [ x - 1(4x) ]
Here we should note that this relation gives two results. first considering (+)ve sign and second considering (-)ve sign.
..................................................................
Considering (+) sign

sec A + tan A = [ x + 1(4x)] + [ x - 1(4x) ]
=> sec A + tan A = [ x + 1(4x)] + [x - 1(4x) ],
On simplification we get -
=> sec A + tan A = 2x .............................. Proved
..........................................................................................
Considering (-) sign

sec A + tan A = [ x + 1(4x)] - [ x - 1(4x) ]
Which on simplification gives -

=> sec A + tan A = 1 / (2x) .............................. Proved

Answer 3

If sec Q = x+1/(4x) = (4x^2 +1)/4x. then
tan Q = sqrt[(4x^2 +1)^2-(4x)^2]/(4x)
So sec Q + tan Q = +/- (4x^2-1)/ 4x+(4x^2 +1)/4x
= 2x or 1/(2x)

Answer 4

sec θ = x + 1/4x
tan^2 θ = sec^2 θ - 1
= (x+ 1/4x)^2 - 1
= x^2 + 1/16x^2 +1/2 - 1
= x^2 + 1/16x^2 - 1/2
= (x - 1/4x)^2
=> tan θ = ± (x - 1/4x)
=>secθ+ tan θ = 2x, or 1/(2x)

Answer 5

draw a traingle , use any theta you like , between any two segments , with your definition
sec(theta) = x + 1/4x

then use geometry to sovle for the drawn triangle

Answer 6

Don't know see photo