# What is the angle that verifies identity tan^2x=6sec x-10?

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We need to find the angle that satisfies: (tan x)^2 = 6*sec x - 10

(tan x)^2 = 6*sec x - 10

=>(sin x)^2 / (cos x)^2 = 6 / cos x - 10

=> [(1 - (cos x)^2)/(cos x)^2] = 6 / cos x - 10

let y = cos x

=> (1 - y^2)/y^2 = 6 / y - 10

=> (1 - y^2) = 6y - 10y^2

=> 9y^2 - 6y + 1 = 0

=> 9y^2 - 3y - 3y + 1 = 0

=> 3y(3y - 1) - 1(3y - 1) = 0

=> (3y - 1)^2 = 0

=> 3y - 1 = 0

=> y = 1/3

cos x = 1/3

=> x = arc cos (1/3)

=> x = 70.52 + n*360 degrees

**The required angle is x = 70.52 + n*360 degrees**

We'll substitute (tan x)^2 = (sec x)^2 - 1

We'll re-write the equation moving all terms to the left side:

(sec x)^2 - 1 - 6sec x + 10 = 0

We'll combine like terms:

(sec x)^2 - 6sec x + 9 = 0

We notice that the expression is a perfect square:

(sec x - 3)^2 = 0

We'll put sec x - 3 = 0

sec x = 3

But sec x = 1/cos x => 1/cos x = 3

cos x = 1/3

Since the value of cosine is in the interval [-1 ; 1], that means that it doesexist an angle that verifies the identity.

We'll determine the angle using inverse trigonometric function.

x = +/- arccos (1/3) + 2kpi

**All the angles x that verify the equation are {+/- arccos (1/3) + 2kpi}.**