To determine the value of the expression, you must know the following trigonometric values:

\(\displaystyle cos(45^{\circ}) = \frac{\sqrt{2}}{2}\)

\(\displaystyle sin (30^{\circ})=\frac{1}{2}\)

Replacing these values, we get:

\(\displaystyle cos^{2}(45^{\circ})-sin(30^{\circ})\)

\(\displaystyle (\frac{\sqrt{2}}{2})^{2}-\frac{1}{2}\)

\(\displaystyle \frac{2}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0\)

\(\displaystyle tan (x)*sin(x) + sec(x)*(cos x)^{2} =\)

\(\displaystyle \frac{sin (x)}{cos (x)}* sin(x) + \frac{1}{cos (x)}*(cos x)^{2} =\)

\(\displaystyle \frac{(sin x)^{2} + (cos x)^{2}}{cos (x)} = \frac{1}{cos (x)} = sec (x)\)

We need to use the following identities:

\(\displaystyle cos(\frac{3\pi}{2}+x)=sin\ x\)

\(\displaystyle sin (\pi+x)=-sin\ x\)

\(\displaystyle cos(\frac{\pi}{2}+x)=-sin\ x\)

\(\displaystyle sin(\pi-x)=sin\ x\)

Use these to simplify the expression as follows:

\(\displaystyle A=cos(\frac{3\pi}{2}+x)+sin (\pi+x)+cos(\frac{\pi}{2}+x)+sin(\pi-x)\)

\(\displaystyle \Rightarrow A=sin\ x-sin\ x-sin\ x+sin\ x = 0\)

\(\displaystyle sin(\frac{\pi}{2})=1\)

\(\displaystyle cos(\frac{\pi}{2})=0\)

Plug these values in:

\(\displaystyle A=\frac{2sin(\frac{\pi}{2})+4cos(\frac{\pi}{2})}{4sin(\frac{\pi}{2})-cos(\frac{\pi}{2})}=\frac{2\times 1+4\times 0}{4\times 1-0}=\frac{2}{4}=\frac{1}{2}=0.5\)

\(\displaystyle tan\ x=\frac{1}{4}\Rightarrow \frac{sin\ x}{cos\ x}=\frac{1}{4}\Rightarrow cos\ x=4sin\ x\)

Substitute \(\displaystyle cos\ x=4sin \ x\) into the expression:

\(\displaystyle D=\frac{sin\ x}{sin\ x-cos\ x}+\frac{sin\ x+cos\ x}{cos\ x}=\frac{sin\ x}{sin\ x-4sin\ x}+\frac{sin\ x+4sin\ x}{4sin\ x}\)

\(\displaystyle \Rightarrow D=\frac{sin\ x}{-3sin\ x}+\frac{5sin\ x}{4sin\ x}\)

\(\displaystyle \Rightarrow D=-\frac{1}{3}+\frac{5}{4}=\frac{-4+15}{12}=\frac{11}{12}\)

 \(\displaystyle cot\ x=3\Rightarrow \frac{cos\ x}{sin\ x}=3\Rightarrow cos\ x= 3sin\ x\)

Now substitute \(\displaystyle cos\ x=3sin\ x\) into the expression:

\(\displaystyle D=\frac{cos^2x}{(sin\ x)(cos\ x)}-\frac{sin\ x+cos\ x}{sin\ x}=\frac{(3sin\ x)^2}{(sin\ x)(3sin\ x)}-\frac{sin\ x+3sin\ x}{sin\ x}\)

\(\displaystyle \Rightarrow D=\frac{9sin^2x}{3sin^2x}-\frac{4sin\ x}{sin\ x}\)

\(\displaystyle \Rightarrow D=\frac{9}{3}-4=3-4=-1\)

We need to use the following identitities:

\(\displaystyle \frac{1}{cos ^2x}=1+tan^2x\)

\(\displaystyle \frac{1}{sin^2x}=1+cot^2x\)

Now substitute them into the expression:

\(\displaystyle A=\frac{sin^2x}{1+tan^2x}-\frac{cos^2x}{1+cot^2x}+1=\frac{sin^2x}{\frac{1}{cos^2x}}-\frac{cos^2x}{\frac{1}{sin^2x}}+1\)

\(\displaystyle \Rightarrow A=(sin^2x)(cos^2x)-(sin^2x)(cos^2x)+1=0+1=1\)

Based on the double angle formula we have, \(\displaystyle sin\ 2x=2(sin\ x)(cos\ x)\).

\(\displaystyle \frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}=\frac{sin\ x+2(sin\ x)(cos\ x)}{sin\ x-2(sin\ x)(cos\ x)}\)

\(\displaystyle =\frac{sinx(1+2cos\ x )}{sinx(1-2cos\ x )}=\frac{1+2cos\ x}{1-2cos\ x}\)

\(\displaystyle =\frac{1+2(\frac{1}{3})}{1-2(\frac{1}{3})}=\frac{\frac{3+2}{3}}{\frac{3-2}{3}}=5\)

\(\displaystyle x=\frac{\pi}{3}\Rightarrow sin\ x=sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\)

Now we can write:

\(\displaystyle x=\frac{\pi}{3}\Rightarrow 2x=2(\frac{\pi}{3})=\frac{2\pi}{3}\Rightarrow cos\ 2x=cos(\frac{2\pi}{3})\)

\(\displaystyle =cos(\pi-\frac{\pi}{3})\)

\(\displaystyle =-cos (\frac{\pi}{3})\)

\(\displaystyle =-\frac{1}{2}\)

Now we can substitute the values:

\(\displaystyle \frac{sin\ x}{sin\ x+cos\ 2x}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\times \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}=\frac{3+\sqrt{3}}{2}\)

\(\displaystyle 326^{\circ}=360^{\circ}-34^{\circ}\Rightarrow sin\ 326^{\circ}=sin(360^{\circ}-34^{\circ})\Rightarrow sin\ 326^{\circ}=-sin\ 34^{\circ}\)

\(\displaystyle 56^{\circ}=90^{\circ}-34^{\circ}\Rightarrow sin\ 56^{\circ}=sin(90^{\circ}-34^{\circ})\Rightarrow sin\ 56^{\circ}=cos\ 34^{\circ}\)

\(\displaystyle 304^{\circ}=270^{\circ}+34^{\circ}\Rightarrow cos\ 304^{\circ}=cos(270^{\circ}+34^{\circ})\Rightarrow cos\ 304^{\circ}=sin\ 34^{\circ}\)

Now we can simplify the expression as follows:

\(\displaystyle \frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}=\frac{-2sin\ 34^{\circ}+3cos\ 34^{\circ}}{sin\ 34^{\circ}}\)

\(\displaystyle =-2+3cot\ 34^{\circ}=-2+3\times 1.5=-2+4.5=2.5\)