# Thread: Worked Solution does not explain why tan x = - √15 is rejected

1. ## Worked Solution does not explain why tan x = - √15 is rejected

Can any kind soul help me in the below?

1. If x is obtuse and 2 tan^2 x = 5 sec x +10, find the value of tan x without using calculator.

obtuse : 90deg < x< 180 deg
2 tan^2 x = 5 sec x + 10
2 (sec^2 x -1) = 5 sec x + 10
2 (sec^2 x -1) - 5 sec x - 10 = 0
2 sec^2 x -2 - 5 sec x - 10 = 0
(2 sec x + 3)(sec x - 4) = 0
sec x = 3/2 or sec x = 4
sec x^2 = (3/2)^2 or sec x^2 = 4^2
sec x^2 = 9/4 or sec x^2 = 16
(sec x^2) - 1 = 9/4 -1 or (sec x^2) - 1 = 16 - 1
tan x^2 = 5/4 or tan x^2 = 15
tan x = √(5/4) or tan x = √15, -√15
tan x = (√5)/2, -(√5)/2 or tan x = √15, -√15
For x to be obtuse, tan x must be in 2nd quadrant => negative
tan x = -(√5)/2 or tan x = -√15 (rejected)

My question: why is tan x = - √15 rejected?

2. Originally Posted by ppppp77

My question: why is tan x = - √15 rejected?

because...

Originally Posted by ppppp77

obtuse : 90deg < x< 180 deg

3. oh my, pickslides, you are awesome!!

4. but can i check, cause my understanding is as below:
tan x = - √15, so x is in 2nd or 4th quadrant
acute angle = 75.5
x = 180 - 75.5 = 104.4 which qualifies as answer because 90deg < x< 180 deg??

5. Hello ppppp77
Originally Posted by ppppp77
but can i check, cause my understanding is as below:
tan x = - √15, so x is in 2nd or 4th quadrant
acute angle = 75.5
x = 180 - 75.5 = 104.4 which qualifies as answer because 90deg < x< 180 deg??
$\tan^2x = 15$
and you then said (correctly):
$\tan x = \pm \sqrt{15}$
You then (correctly) rejected $\tan x = \sqrt{15}$ because $x$ is obtuse and therefore $\tan x< 0$.

But does that mean that $\tan x = -\sqrt{15}$ is necessarily correct? No, before you can say that, there's one more test to apply, and that is to see whether this value satisfies the original equation; which was:
$2\tan^2x = 5 \sec x +10,\; 90^o
Now the LHS is $2\times15=30$. But what about $\sec x$ on the RHS? Well, using $\sec^2x = 1+ \tan^2 x$, we get:
$\sec^2x=16$

$\Rightarrow \sec x = \pm4$
and, of course, for $90^o, we must take the negative sign, giving
$\sec x = -4$
and the RHS then is $-20+10 = -10$, not $+20+10= 30$ (which is what we wanted).

So $\tan x = -\sqrt{15}$ doesn't work either.

Tricky little beast, isn't it?