12 (a) Show that the equation $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ can be written in the form a \csc^2 x + b \csc x + c = 0 12 (b) Hence, given $x$ is obtuse and $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ find the exact value of $\tan x$ - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 1

Given that $x$ is obtuse, we can analyze the quadratic equation obtained in part (a):

$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$

From our previous work, we found:

$\csc x = \frac{3}{2} \, \text{or} \, \csc x = -1$

The valid solution must satisfy $|\csc x| \geq 1$, therefore, we take:

$\csc x = \frac{3}{2}$

The definition of cosecant is given by:

$\csc x = \frac{1}{\sin x}$

Thus:

$\sin x = \frac{2}{3}$

Now, using the identity:

$\tan x = \frac{\sin x}{\cos x}$

And recalling that for an obtuse angle, we have:

$\cos x = -\sqrt{1 - \sin^2 x}$

Calculating this gives:

$\cos x = -\sqrt{1 - \left(\frac{2}{3}\right)^2} = -\sqrt{1 - \frac{4}{9}} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$

Now substituting back into the tangent formula:

$\tan x = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}$

Thus, the exact value of $\tan x$ is:

$\tan x = -\frac{2\sqrt{5}}{5}$