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Hint: Try to simplify the left-hand side of the equation that we need to prove by substituting x and y with their respective values as given in the question. Finally, use the formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to reach the desired result.

Complete step-by-step answer:

Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.

First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $

We will now solve the left-hand side of the equation given in the question.

${{\left( \dfrac{x-y}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}$

Now we will substitute x and y as given in the question. On doing so, we get

${{\left( \dfrac{\cot A+\cos A-\left( \cot A-\cos A \right)}{\cot A+\cos

A+\operatorname{cotA}-cosA} \right)}^{2}}+{{\left( \dfrac{\cot A+\cos A-\left( \cot A-\cos A

\right)}{2} \right)}^{2}}$

$={{\left( \dfrac{\cot A+\cos A-\cot A+\cos A}{\cot A+\cos A+\operatorname{cotA}-cosA}

\right)}^{2}}+{{\left( \dfrac{\cot A+\cos A-\cot A+\cos A}{2} \right)}^{2}}$

$={{\left( \dfrac{\cos A}{\cot A} \right)}^{2}}+{{\left( \dfrac{\cos A}{{}} \right)}^{2}}$

$={{\left( \dfrac{\cos A}{\cot A} \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

Now we know that $\cot A=\dfrac{\cos A}{\sin A}$ . So, our equation becomes

${{\left( \dfrac{\sin A\cos A}{\cos A} \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

$={{\left( \sin A \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

We know that the sum of the squares of the sine and cosine function of an angle is equal to

1. So, our expression turns out to be:

${{\left( \sin A \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

$=1$

As we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation in the question. Therefore, we can say that we have proved that ${{\left( \dfrac{x-y}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}=1$ .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, it would help if you remember the properties related to complementary angles and trigonometric ratios.

Complete step-by-step answer:

Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.

First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $

We will now solve the left-hand side of the equation given in the question.

${{\left( \dfrac{x-y}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}$

Now we will substitute x and y as given in the question. On doing so, we get

${{\left( \dfrac{\cot A+\cos A-\left( \cot A-\cos A \right)}{\cot A+\cos

A+\operatorname{cotA}-cosA} \right)}^{2}}+{{\left( \dfrac{\cot A+\cos A-\left( \cot A-\cos A

\right)}{2} \right)}^{2}}$

$={{\left( \dfrac{\cot A+\cos A-\cot A+\cos A}{\cot A+\cos A+\operatorname{cotA}-cosA}

\right)}^{2}}+{{\left( \dfrac{\cot A+\cos A-\cot A+\cos A}{2} \right)}^{2}}$

$={{\left( \dfrac{\cos A}{\cot A} \right)}^{2}}+{{\left( \dfrac{\cos A}{{}} \right)}^{2}}$

$={{\left( \dfrac{\cos A}{\cot A} \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

Now we know that $\cot A=\dfrac{\cos A}{\sin A}$ . So, our equation becomes

${{\left( \dfrac{\sin A\cos A}{\cos A} \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

$={{\left( \sin A \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

We know that the sum of the squares of the sine and cosine function of an angle is equal to

1. So, our expression turns out to be:

${{\left( \sin A \right)}^{2}}+{{\left( \cos A \right)}^{2}}$

$=1$

As we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation in the question. Therefore, we can say that we have proved that ${{\left( \dfrac{x-y}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}=1$ .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, it would help if you remember the properties related to complementary angles and trigonometric ratios.

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