Question

If \[x=\operatorname{cosecA}+cosA\]and \[y=\operatorname{cosecA}-cosA\], then prove that

\[{{\left( \dfrac{2}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}-1=0\].

Answer 1

Hint: In this question, we first need to find the values of the sum of x and y and the difference of x and y. Then substitute these values in the given expression and simplify it further using the trigonometric identities. Then add the respective terms and then again use the trigonometric identity to get the result.

\[\sin \theta =\dfrac{1}{cosec\theta }\]

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Complete step-by-step answer:


Now, from the given conditions in the question we have

\[x=\operatorname{cosecA}+cosA\]

\[y=\operatorname{cosecA}-cosA\]

Let us first find the sum of x and y from the above conditions

\[\Rightarrow x+y\]

Now, on substituting the respective values we get,

\[\Rightarrow \operatorname{cosecA}+cosA+\operatorname{cosecA}-cosA\]

Now, on further simplification we get,

\[\Rightarrow 2\operatorname{cosecA}\]

\[\therefore x+y=2\operatorname{cosecA}\]

Let us now find the difference of x and y from the given conditions

\[\Rightarrow x-y\]

Now, on substituting the respective values we get,

\[\Rightarrow \operatorname{cosecA}+cosA-\left( \operatorname{cosecA}-cosA \right)\]

Now, this can also be written as

\[\Rightarrow \operatorname{cosecA}+cosA-\operatorname{cosecA}+cosA\]

Now, on further simplification we get,

\[\Rightarrow 2cosA\]

\[\therefore x-y=2cosA\]

Now, from the given equation in the question we have

\[\Rightarrow {{\left( \dfrac{2}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}-1=0\]

Let us now consider the left hand side in the above given equation

\[\Rightarrow {{\left( \dfrac{2}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}-1\]

Now, let us substitute the respective values of sum and difference of x and y

\[\Rightarrow {{\left( \dfrac{2}{2\operatorname{cosecA}} \right)}^{2}}+{{\left( \dfrac{2cosA}{2}

\right)}^{2}}-1\]

Now, on cancelling out the common terms we get,

\[\Rightarrow {{\left( \dfrac{1}{\operatorname{cosecA}} \right)}^{2}}+{{\left( \dfrac{cosA}{1}

\right)}^{2}}-1\]

As we already know that from the trigonometric identities the relation between sine and

cosecant function can be expressed as

\[\sin \theta =\dfrac{1}{cosec\theta }\]

Now, by using the above trigonometric identity the expression can be further written as

\[\Rightarrow {{\left( sinA \right)}^{2}}+{{\left( cosA \right)}^{2}}-1\]

Now, this can be further written as

\[\Rightarrow si{{n}^{2}}A+co{{s}^{2}}A-1\]

As we already know from the trigonometric identities that

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Now, the above expression can be further written as

\[\Rightarrow 1-1\]

\[\Rightarrow 0\]

Hence, \[{{\left( \dfrac{2}{x+y} \right)}^{2}}+{{\left( \dfrac{x-y}{2} \right)}^{2}}-1=0\]


Note:
Instead of finding the sum and difference separately and then substituting we can directly substitute the values of x and y and then simplify it further accordingly. Both the methods give the same results.

It is important to note that we need to use the corresponding trigonometric identities to solve the obtained expressions further. Because considering the wrong identity or neglecting any of the terms changes the result completely.