Show that $\text{cot}2\theta =\frac{1-\text{ta}{\text{n}}^2\theta }{2\text{tan}\theta }$.

Verify that $x=\text{tan}\theta$ and $x=-\text{cot}\theta$ satisfy the equation $x^2+\left(2\text{cot}2\theta \right)x-1=0$.

Hence, or otherwise, show that the exact value of $\text{tan}\frac{\pi }{12}=2-\sqrt{3}$.

Using the results from parts (b) and (c) find the exact value of $\text{tan}\frac{\pi }{24}-\text{cot}\frac{\pi }{24}$.

Give your answer in the form $a+b\sqrt{3}$ where $a$, $b\in \mathbb{Z}$.

stating the relationship between $\cot$ and $\tan$ and stating the identity for $\text{tan}2\theta$       M1

$\text{cot}2\theta =\frac{1}{\text{tan}2\theta }$ and $\text{tan}2\theta =\frac{2\text{tan}\theta }{1-\text{ta}{\text{n}}^2\theta }$

⇒ $\text{cot}2\theta =\frac{1-\text{ta}{\text{n}}^2\theta }{2\text{tan}\theta }$     AG

[1 mark]

METHOD 1

attempting to substitute $\text{tan}\theta$ for $x$ and using the result from (a)      M1

LHS = $\text{ta}{\text{n}}^2\theta +2\text{tan}\theta \left(\frac{1-\text{ta}{\text{n}}^2\theta }{2\text{tan}\theta }\right)-1$      A1

$\text{ta}{\text{n}}^2\theta +1-\text{ta}{\text{n}}^2\theta -1=0$(= RHS)      A1

so $x=\text{tan}\theta$ satisfies the equation      AG

attempting to substitute $-\text{cot}\theta$ for $x$ and using the result from (a)       M1

LHS = $\text{co}{\text{t}}^2\theta -2\text{cot}\theta \left(\frac{1-\text{ta}{\text{n}}^2\theta }{2\text{tan}\theta }\right)-1$      A1

$=\frac{1}{\text{ta}{\text{n}}^2\theta }-\left(\frac{1-\text{ta}{\text{n}}^2\theta }{\text{ta}{\text{n}}^2\theta }\right)-1$      A1

$\frac{1}{\text{ta}{\text{n}}^2\theta }-\frac{1}{\text{ta}{\text{n}}^2\theta }+1-1=0$(= RHS)     A1

so $x=-\text{cot}\theta$ satisfies the equation      AG

METHOD 2

let $\alpha =\text{tan}\theta$ and $\beta =-\text{cot}\theta$

attempting to find the sum of roots       M1

$\alpha +\beta =\text{tan}\theta -\frac{1}{\text{tan}\theta }$

         $=\frac{\text{ta}{\text{n}}^2\theta -1}{\text{tan}\theta }$     A1

         $=-2\text{cot}2\theta$ (from part (a))     A1

attempting to find the product of roots         M1

$\alpha \beta =\text{tan}\theta \times \left(-\text{cot}\theta \right)$     A1

= −1     A1

the coefficient of $x$ and the constant term in the quadratic are $2\text{cot}2\theta$ and −1 respectively        R1

hence the two roots are $\alpha =\text{tan}\theta$ and $\beta =-\text{cot}\theta$       AG

[7 marks]

METHOD 1

$x=\text{tan}\frac{\pi }{12}$ and $x=-\text{cot}\frac{\pi }{12}$ are roots of $x^2+\left(2\text{cot}\frac{\pi }{6}\right)x-1=0$        R1

Note: Award R1 if only $x=\text{tan}\frac{\pi }{12}$ is stated as a root of $x^2+\left(2\text{cot}\frac{\pi }{6}\right)x-1=0$.

$x^2+2\sqrt{3}x-1=0$        A1

attempting to solve their quadratic equation         M1

$x=-\sqrt{3}\pm 2$        A1

$\text{tan}\frac{\pi }{12}>0$  ()        R1

so $\text{tan}\frac{\pi }{12}=2-\sqrt{3}$      AG

METHOD 2

attempting to substitute $\theta =\frac{\pi }{12}$ into the identity for $\text{tan}2\theta$           M1

$\text{tan}\frac{\pi }{6}=\frac{2\text{tan}\frac{\pi }{12}}{1-\text{ta}{\text{n}}^2\frac{\pi }{12}}$

$\text{ta}{\text{n}}^2\frac{\pi }{12}+2\sqrt{3}\text{tan}\frac{\pi }{12}-1=0$     A1

attempting to solve their quadratic equation      M1

$\text{tan}\frac{\pi }{12}=-\sqrt{3}\pm 2$     A1

$\text{tan}\frac{\pi }{12}>0$      R1

so $\text{tan}\frac{\pi }{12}=2-\sqrt{3}$      AG

[5 marks]

$\text{tan}\frac{\pi }{24}-\text{cot}\frac{\pi }{24}$ is the sum of the roots of $x^2+\left(2\text{cot}\frac{\pi }{12}\right)x-1=0$        R1

$\text{tan}\frac{\pi }{24}-\text{cot}\frac{\pi }{24}=-2\text{cot}\frac{\pi }{12}$      A1

$=\frac{-2}{2-\sqrt{3}}$      A1

attempting to rationalise their denominator       (M1)

$=-4-2\sqrt{3}$       A1A1

[6 marks]