Question

How do you solve tan x + sec x = 2?

Answer 1

Given equation

`secx+tanx=2........[1]`

Again we know

`sec^2x-tan^2x=1.......[2]`

Dividing [2] by [1]

`secx-tanx=1/2.......[3]`

Adding [1] and [3] we get

`2secx=2+1/2=5/2`

`=>secx=5/4`

`=>cosx=4/5=cosalpha`,where `alpha=cos^-1(4/5)`

So `x=2npi±alpha" where "n in ZZ`

Again subtracting [3] from [1] we get a solution in another form

`2tanx=3/2`

`tanx=3/4=tanbeta`,where `beta=tan^-1(3/4)`

So `x=npi+beta" where "n in ZZ`

Alternatve

Given equation

`secx+tanx=2`

`=>1/cosx+sinx/cosx=2`

`=>(sqrt(1+sinx))^2/sqrt((1+sinx)(1-sinx))=2`
[`1+sinx!=0`]

`=>(sqrt(1+sinx))/sqrt(1-sinx)=2`

`=>(1+sinx)/(1-sinx)=4`

`=>(1+sinx)=4(1-sinx)`

`=>5sinx=3=>sinx=3/5=sinsin^-1(3/5)`

`=>x=npi+(-1)^nsin^-1(3/5)`

Answer 2

`x = arcsin(3/5) + 2pin`.

Explanation:

Here's another way of solving this problem.

We know that `tantheta = sintheta/costheta` and `sectheta = 1/costheta`. So,

`sinx/cosx + 1/cosx= 2`

`(sinx + 1)/cosx= 2`

`sinx + 1 = 2cosx`

Square both sides.

`(sinx + 1)^2 = (2cosx)^2`

`sin^2x+ 2sinx + 1 = 4cos^2x`

We use the identity `sin^2theta + cos^2theta = 1-> cos^2theta = 1 - sin^2theta`.

`sin^2x + 2sinx + 1 = 4(1 - sin^2x)`

`sin^2x+ 2sinx + 1 = 4 - 4sin^2x`

`5sin^2x + 2sinx - 3 = 0`

We let `t = sinx`.

`5t^2 + 2t - 3 = 0`

`5t^2 + 5t - 3t - 3 = 0`

`5t(t + 1) - 3(t + 1) = 0`

`(5t - 3)(t + 1)= 0`

`t = 3/5 and -1`

`sinx = 3/5 and sinx = -1`

`x = pi - arcsin(3/5) + 2pin, arcsin(3/5) + 2pin, (3pi)/2 + 2pin`

However, `pi - arcsin(3/5) + 2pin` and `(3pi)/2 + 2pin` extraneous. `pi - arcsin(3/5)` since tangent is negative in quadrant II and so is cosine. `(3pi)/2` is extraneous since it is an asymptote in the secant function and in the tangent function.

Therefore, the only actual solution is `x = arcsin(3/5) + 2pin`, where `n` is an integer.

Hopefully this helps!