Find an expression for $g \circ f(x)$, stating its domain.

Hence show that $g \circ f(x) = \frac{{\sin x + \cos x}}{{\sin x - \cos x}}$.

Let $y = g \circ f(x)$, find an exact value for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ at the point on the graph of $y = g \circ f(x)$ where $x = \frac{\pi }{6}$, expressing your answer in the form $a + b\sqrt 3 ,{\text{ }}a,{\text{ }}b \in \mathbb{Z}$.

Show that the area bounded by the graph of $y = g \circ f(x)$, the $x$-axis and the lines $x = 0$ and $x = \frac{\pi }{6}$ is $\ln \left( {1 + \sqrt 3 } \right)$.

$g \circ f(x) = \frac{{\tan x + 1}}{{\tan x - 1}}$     A1

$x \ne \frac{\pi }{4},{\text{ }}0 \le x < \frac{\pi }{2}$     A1

[2 marks]

$\frac{{\tan x + 1}}{{\tan x - 1}} = \frac{{\frac{{\sin x}}{{\cos x}} + 1}}{{\frac{{\sin x}}{{\cos x}} - 1}}$     M1A1

$= \frac{{\sin x + \cos x}}{{\sin x - \cos x}}$     AG

[2 marks]

METHOD 1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x - \cos x)}^2}}}$     M1(A1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(2\sin x\cos x - {{\cos }^2}x - {{\sin }^2}x) - (2\sin x\cos x + {{\cos }^2}x + {{\sin }^2}x)}}{{{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x}}$

$= \frac{{ - 2}}{{1 - \sin 2x}}$

Substitute $\frac{\pi }{6}$ into any formula for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1

$\frac{{ - 2}}{{1 - \sin \frac{\pi }{3}}}$

$= \frac{{ - 2}}{{1 - \frac{{\sqrt 3 }}{2}}}$     A1

$= \frac{{ - 4}}{{2 - \sqrt 3 }}$

$= \frac{{ - 4}}{{2 - \sqrt 3 }}\left( {\frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}} \right)$     M1

$= \frac{{ - 8 - 4\sqrt 3 }}{1} =  - 8 - 4\sqrt 3$     A1

METHOD 2

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\tan x - 1){{\sec }^2}x - (\tan x + 1){{\sec }^2}x}}{{{{(\tan x - 1)}^2}}}$     M1A1

$= \frac{{ - 2{{\sec }^2}x}}{{{{(\tan x - 1)}^2}}}$     A1

$= \frac{{ - 2{{\sec }^2}\frac{\pi }{6}}}{{{{\left( {\tan \frac{\pi }{6} - 1} \right)}^2}}} = \frac{{ - 2\left( {\frac{4}{3}} \right)}}{{{{\left( {\frac{1}{{\sqrt 3 }} - 1} \right)}^2}}} = \frac{{ - 8}}{{{{\left( {1 - \sqrt 3 } \right)}^2}}}$     M1

Note: Award M1 for substitution $\frac{\pi }{6}$.

$\frac{{ - 8}}{{{{\left( {1 - \sqrt 3 } \right)}^2}}} = \frac{{ - 8}}{{\left( {4 - 2\sqrt 3 } \right)}}\frac{{\left( {4 + 2\sqrt 3 } \right)}}{{\left( {4 + 2\sqrt 3 } \right)}} =  - 8 - 4\sqrt 3$     M1A1

[6 marks]

Area $\left| {\int_0^{\frac{\pi }{6}} {\frac{{\sin x + \cos x}}{{\sin x - \cos x}}{\text{d}}x} } \right|$     M1

$= \left| {\left[ {\ln \left| {\sin x - \cos x} \right|} \right]_0^{\frac{\pi }{6}}} \right|$     A1

Note:     Condone absence of limits and absence of modulus signs at this stage.

$= \left| {\ln \left| {\sin \frac{\pi }{6} - \cos \frac{\pi }{6}} \right| - \ln \left| {\sin 0 - \cos 0} \right|} \right|$     M1

$= \left| {\ln \left| {\frac{1}{2} - \frac{{\sqrt 3 }}{2}} \right| - 0} \right|$

$= \left| {\ln \left( {\frac{{\sqrt 3  - 1}}{2}} \right)} \right|$     A1

$=  - \ln \left( {\frac{{\sqrt 3  - 1}}{2}} \right) = \ln \left( {\frac{2}{{\sqrt 3  - 1}}} \right)$     A1

$= \ln \left( {\frac{2}{{\sqrt 3  - 1}} \times \frac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}}} \right)$     M1

$= \ln \left( {\sqrt 3  + 1} \right)$     AG

[6 marks]

Total [16 marks]