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 Mechanics - Question 12 - 2019 - Paper 1

Given that x is obtuse, we know that ( \csc x \geq 1 ), and that thus:

From our earlier equation:

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

We can solve this quadratic equation for ( \csc x ) using the quadratic formula:

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

Substituting in the values:

$\csc x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} = \frac{4 \pm \sqrt{16 + 48}}{8} = \frac{4 \pm \sqrt{64}}{8} = \frac{4 \pm 8}{8}$

This gives:

$\csc x = \frac{12}{8} = \frac{3}{2} \quad \text{or} \quad \csc x = \frac{-4}{8} = -\frac{1}{2}\, \text{(reject as } |csc x| \ge 1 \text{) }$

Thus, ( \csc x = \frac{3}{2} ). Therefore, the value of ( \sin x = \frac{2}{3} ) and since x is obtuse, we find:

$\tan x = \frac{\sin x}{\cos x} \quad \text{where } \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} \implies \cos x = -\frac{\sqrt{5}}{3}$

Thus:

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