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??

In your working you had (correctly):$\displaystyle \tan^2x = 15$

and you then said (correctly):$\displaystyle \tan x = \pm \sqrt{15}$

You then (correctly) rejected $\displaystyle \tan x = \sqrt{15}$ because $\displaystyle x$ is obtuse and therefore $\displaystyle \tan x< 0$.

But does that mean that $\displaystyle \tan x = -\sqrt{15}$ 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: $\displaystyle 2\tan^2x = 5 \sec x +10,\; 90^o<x<180^o$

Now the LHS is $\displaystyle 2\times15=30$. But what about $\displaystyle \sec x$ on the RHS? Well, using $\displaystyle \sec^2x = 1+ \tan^2 x$, we get:$\displaystyle \sec^2x=16$

$\displaystyle \Rightarrow \sec x = \pm4$

and, of course, for $\displaystyle 90^o<x<180^o$, we must take the negative sign, giving $\displaystyle \sec x = -4$

and the RHS then is $\displaystyle -20+10 = -10$, not $\displaystyle +20+10= 30$ (which is what we wanted).

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

Tricky little beast, isn't it?

Grandad