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]