Question

if sec theta = x + 1/4x ,prove that sec theta +tan theta = 2x or 1/2x

Answer 1

secA = x + 1/4x = (4x^2 + 1)/4x⇒x = 4x and c = 4x^2 + 1

                                                    c^2 = x^2 + y^2

                                                  y= sqr (  c^2 - x^2)

                                                 y = sqrt [(4x^2 + 1)^2 - (4x)^2]

                                                y = sqrt ( 16x^4 + 8x^2 + 1 - 16x^2)

                                                y = sqrt ( 16x^4 -8x^2 + 1+

                                                y = sqrt [ ( 4x^2 - 1)^2]

                                               y = 4x^2 - 1

tanA = ( 4x^2 - 1)/4x

sec A + tan A = (4x^2 + 1)/4x +( 4x^2 - 1)/4x =

                          8x^2/(4x) = 2x

I wrote A instead teta

Answer 2

* 1 + tan^2 theta = sec^2 theta

sec^2 theta - tan^2 theta = 1

(sec theta + tan theta) (sec theta - tan theta) = 1     [since a^2 - b^2 = (a+b) (a-b) ]

{the product of two no.s is 1 that means they must be reciprocal of each other}

this implies that (sec theta + tan theta) and (sec theta - tan theta) are reciprocals of each other.

so let (sec theta + tan theta) = p ...............................eq. 1

and therefore (sec theta - tan theta) = 1/p...................eq.2

adding eq.1 and eq.2 we get,

2 sec theta = p +1/p

therefore , sec theta = p/2 + 1/ 2p

x + 1/4x = p/2 + 1/2p                                 {since, sec theta = x + 1/4x (given)}

on compairing....

either p/2 = x or p/2 = 1/4x

if p/2 = x then,

p = 2x

this implies that sec theta + tan theta = 2x                   {since sec theta + tan theta = p}

or if p/2=1/4x then,

p = 1/2x

this implies that sec theta = tan theta = 1/2x

Answer 3

sec theta = x + 1/4x

sec^2 theta = x^2 + 1/16x^2 + 1/2

sec^2 theta -1 = x^2  + 1/16x^2 - 1/2

So, tan^2 theta = x^2 + 1/16x^2 - 1/2 which is (x - 1/4x)^2

so, tan theta = x - 1/4x

then, sec theta + tan theta = x + 1/4x + x - 1/4x = 2x