Since \(\displaystyle 52^{\circ}=2\times 26^{\circ}\), we can use a double angle formula:

\(\displaystyle cos\ 2x=1-2sin^2x\Rightarrow cos\ 52^{\circ}=1-2sin^2\ (26)^{\circ}\)

Substituting \(\displaystyle sin \26^{\circ}=t\), we get \(\displaystyle cos\ 52^{\circ}=1-2sin^2\ (26)^{\circ}=1-2t^2\).

To solve the problem, you need to know the following information:

\(\displaystyle sin(60^{\circ}) = \frac{\sqrt{3}}{2}\)

\(\displaystyle tan(45^{\circ})=1\)

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

Replace the trigonometric functions with these values:

\(\displaystyle \frac{4sin^2(60^{\circ})-tan(45^{\circ})}{3sin(45^{\circ})}\)

\(\displaystyle \frac{4(\frac{\sqrt{3}}{2})^2-(1)}{3(\frac{\sqrt{2}}{2})}\)

\(\displaystyle \frac{4(\frac{3}{4})-(1)}{(\frac{3\sqrt{2}}{2})}\)

\(\displaystyle \frac{2}{(\frac{3\sqrt{2}}{2})}\)

\(\displaystyle \frac{4}{3\sqrt{2}} = \frac{4\cdot \sqrt{2}}{3\sqrt{2}\cdot \sqrt{2}} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}\)

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:

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

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