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

Given that (x) is obtuse and

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

we proceed to find (\csc x):

Using the equation derived in part (a), we set:

$4 \csc^2 x - 4 \csc x - 3 = 0$

Applying the quadratic formula:

$\csc x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}$

Calculate the discriminant:

$\sqrt{16 + 48} = \sqrt{64} = 8$

Hence,

$\csc x = \frac{4 \pm 8}{8}$

This gives two potential solutions:

1. (\csc x = \frac{12}{8} = \frac{3}{2})
2. (\csc x = \frac{-4}{8} = -\frac{1}{2} \text{ (reject as (|csc x| \geq 1|)})

Since we accept (\csc x = \frac{3}{2}), we then find (\sin x):

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

Use the identity for tan:

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

Calculating (\cos x):

Since in triangle terms for obtuse angle,

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

Thus,

$\cos x = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$

Finally, we find:

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

Therefore, the exact value of (\tan x) is:

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