The expressions in $x\And y$ is given as: $x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \sec \left( {{\sin }^{-1}}a \right) \right) \right) \right) \right)$; $y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( {{\cos }^{-1}}a \right) \right) \right) \right) \right)$, where $a\in \left[ 0,1 \right]$. Then which of the following is the correct option:
(a) ${{x}^{2}}+{{a}^{2}}=3$
(b) $x=y$
(c) ${{y}^{2}}+{{a}^{2}}=3$
(d) All of these
Answer
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Hint: To find the correct option, we are going to individually solve $x\And y$. We are going to solve $x\And y$ by taking the inverse trigonometric functions as some angle says $\theta $. For e.g., considering ${{\sin }^{-1}}a$ as $\theta $. Then we will take sine on both the sides and from $\sin \theta $, we can find other trigonometric ratios. Then simplify $x\And y$ using this way.
Complete step by step answer:
In the above problem, we have given the value of $x\And y$ as follows:
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \sec \left( {{\sin }^{-1}}a \right) \right) \right) \right) \right)$;
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( {{\cos }^{-1}}a \right) \right) \right) \right) \right)$
First of all, we are going to solve x by assuming ${{\sin }^{-1}}a=\theta $ then taking sine on both the sides we get,
$\sin \left( {{\sin }^{-1}}a \right)=\sin \theta $
We know that multiplying any term with its inverse will give us 1 so the evaluation of $\sin \left( {{\sin }^{-1}} \right)=1$ and using this relation in the above we get,
$a=\sin \theta $
Now, putting ${{\sin }^{-1}}a=\theta $ in x we get,
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \sec \theta \right) \right) \right) \right)$
We are going to find the value of $\sec \theta $ using $\sin \theta $ which we have calculated above in the following manner:
$a=\sin \theta $
We know that:
$\sec \theta =\dfrac{1}{\cos \theta }$
And we know the trigonometric identity which states that:
$\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
\end{align}$
Taking square root on both the sides of the above equation we get,
$\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Substituting the value of $\sin \theta =a$ in the above equation we get,
$\cos \theta =\sqrt{1-{{a}^{2}}}$
Using the above relation in $\sec \theta =\dfrac{1}{\cos \theta }$ we get,
$\sec \theta =\dfrac{1}{\sqrt{1-{{a}^{2}}}}$
Substituting the above value in new $x$ we get,
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right) \right) \right) \right)$
Now, we are going to assume ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ and then taking $\cot $ on both the sides of this equation we get,
$\begin{align}
& \cot {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\cot \alpha \\
& \Rightarrow \dfrac{1}{\sqrt{1-{{a}^{2}}}}=\cot \alpha \\
\end{align}$
Substituting ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ in new $x$ we get,
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( \alpha \right) \right) \right)$
Now, we are going to find the value of $\cos \alpha $ from $\cot \alpha $ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\cot \alpha $
We know that $\cot \alpha $ is the ratio of base (B) to perpendicular (P) the of the right angled triangle so equating the above equation to $\dfrac{B}{P}$ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\cot \alpha =\dfrac{B}{P}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “B” as 1 and “P” as $\sqrt{1-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{\left( \sqrt{1-{{a}^{2}}} \right)}^{2}}+{{1}^{2}}} \\
& \Rightarrow H=\sqrt{1-{{a}^{2}}+1} \\
& \Rightarrow H=\sqrt{2-{{a}^{2}}} \\
\end{align}$
Now, we know that $\cos \alpha =\dfrac{B}{H}$ so substituting the values of “B and H” in this cosine expression and we get,
$\cos \alpha =\dfrac{1}{\sqrt{2-{{a}^{2}}}}$
Substituting the above value of $\cos \alpha $ in new $x$ and we get,
$x=\csc \left( {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right) \right)$
Now, we are going to assume ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ and then taking $\tan $ on both the sides of this equation we get,
$\begin{align}
& \tan {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta \\
& \Rightarrow \left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta \\
\end{align}$
Substituting ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ in new $x$ we get,
$x=\csc \left( \beta \right)$
Now, we are going to find the value of $\csc \beta $ from $\tan \beta $ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta $
We know that $\tan \beta $ is the ratio of perpendicular (P) to the base (B) of the right angled triangle so equating the above equation to $\dfrac{P}{B}$ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta =\dfrac{P}{B}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “P” as 1 and “B” as $\sqrt{2-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{1}^{2}}+{{\left( \sqrt{2-{{a}^{2}}} \right)}^{2}}} \\
& \Rightarrow H=\sqrt{1+2-{{a}^{2}}} \\
& \Rightarrow H=\sqrt{3-{{a}^{2}}} \\
\end{align}$
Now, we know that $\csc \beta =\dfrac{H}{P}$ so substituting the values of “H and P” in this cosecant expression and we get,
$\csc \beta =\sqrt{3-{{a}^{2}}}$
Using the above relation in new $x$ we get,
\[x=\sqrt{3-{{a}^{2}}}\]
Taking square on both the sides of the above equation we get,
\[\begin{align}
& {{x}^{2}}={{\left( \sqrt{3-{{a}^{2}}} \right)}^{2}} \\
& \Rightarrow {{x}^{2}}=3-{{a}^{2}} \\
\end{align}\]
Adding ${{a}^{2}}$ on both the sides of the above equation we get,
${{x}^{2}}+{{a}^{2}}=3$ …….. (1)
From the above, option (a) is correct.
Similarly, we are going to solve y in the above problem as follows:
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( {{\cos }^{-1}}a \right) \right) \right) \right) \right)$
First of all, we are going to solve x by assuming ${{\cos }^{-1}}a=\theta $ then taking cosine on both the sides we get,
$\cos \left( {{\cos }^{-1}}a \right)=\theta $
We know that multiplying any term with its inverse will give us 1 so the evaluation of $\cos \left( {{\cos }^{-1}} \right)=1$ and using this relation in the above we get,
$a=\cos \theta $
Now, putting ${{\cos }^{-1}}a=\theta $ in y we get,
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( \theta \right) \right) \right) \right) \right)$
We are going to find the value of $\csc \theta $ using $\cos \theta $ which we have calculated above in the following manner:
$a=\cos \theta $
We know that:
$\csc \theta =\dfrac{1}{\sin \theta }$
And we know the trigonometric identity which states that:
$\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& \Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \\
\end{align}$
Taking square root on both the sides of the above equation we get,
$\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$
Substituting the value of $\cos \theta =a$ in the above equation we get,
$\sin \theta =\sqrt{1-{{a}^{2}}}$
Using the above relation in $\csc \theta =\dfrac{1}{\sin \theta }$ we get,
$\csc \theta =\dfrac{1}{\sqrt{1-{{a}^{2}}}}$
Substituting the above value in new y we get,
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right) \right) \right) \right)$
Now, we are going to assume ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ and then taking $\tan $ on both the sides of this equation we get,
$\begin{align}
& \tan {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\tan \alpha \\
& \Rightarrow \dfrac{1}{\sqrt{1-{{a}^{2}}}}=\tan \alpha \\
\end{align}$
Substituting ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ in new y we get,
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( \alpha \right) \right) \right)$
Now, we are going to find the value of $\sin \alpha $ from $\tan \alpha $ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\tan \alpha $
We know that $\tan \alpha $ is the ratio of perpendicular (P) to base (B) of the right angled triangle so equating the above equation to $\dfrac{P}{B}$ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\tan \alpha =\dfrac{P}{B}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “P” as 1 and “B” as $\sqrt{1-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{1}^{2}}+{{\left( \sqrt{1-{{a}^{2}}} \right)}^{2}}} \\
& \Rightarrow H=\sqrt{1+1-{{a}^{2}}} \\
& \Rightarrow H=\sqrt{2-{{a}^{2}}} \\
\end{align}$
Now, we know that $\sin \alpha =\dfrac{P}{H}$ so substituting the values of “P and H” in this cosine expression and we get,
$\sin \alpha =\dfrac{1}{\sqrt{2-{{a}^{2}}}}$
Substituting the above value of $\sin \alpha $ in new y and we get,
$y=\sec \left( {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right) \right)$
Now, we are going to assume ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ and then taking $\cot $ on both the sides of this equation we get,
$\begin{align}
& \cot {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta \\
& \Rightarrow \left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta \\
\end{align}$
Substituting ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ in new $y$ we get,
$y=\sec \left( \beta \right)$
Now, we are going to find the value of $\sec \beta $ from $\cot \beta $ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta $
We know that $\cot \beta $ is the ratio of base (B) to the perpendicular (P) of the right angled triangle so equating the above equation to $\dfrac{B}{P}$ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta =\dfrac{B}{P}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “B” as 1 and “P” as $\sqrt{2-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{\left( \sqrt{2-{{a}^{2}}} \right)}^{2}}+{{1}^{2}}} \\
& \Rightarrow H=\sqrt{2-{{a}^{2}}+1} \\
& \Rightarrow H=\sqrt{3-{{a}^{2}}} \\
\end{align}$
Now, we know that $\sec \beta =\dfrac{H}{B}$ so substituting the values of “H and B” in this secant expression and we get,
$\sec \beta =\sqrt{3-{{a}^{2}}}$
Using the above relation in new $y$ we get,
\[y=\sqrt{3-{{a}^{2}}}\]
Taking square on both the sides of the above equation we get,
\[\begin{align}
& {{y}^{2}}={{\left( \sqrt{3-{{a}^{2}}} \right)}^{2}} \\
& \Rightarrow {{y}^{2}}=3-{{a}^{2}} \\
\end{align}\]
Adding ${{a}^{2}}$ on both the sides of the above equation we get,
\[{{y}^{2}}+{{a}^{2}}=3\]………… (2)
From the above option (c) is correct.
From eq. (1) and eq. (2) we get,
\[\begin{align}
& {{x}^{2}}+{{a}^{2}}=3 \\
& {{y}^{2}}+{{a}^{2}}=3 \\
\end{align}\]
Rearranging the above two equations we get,
\[\begin{align}
& {{x}^{2}}=3-{{a}^{2}} \\
& {{y}^{2}}=3-{{a}^{2}} \\
\end{align}\]
From the above, ${{x}^{2}}={{y}^{2}}$ so subtracting ${{y}^{2}}$ on both the sides of this equation we get,
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=0 \\
& \Rightarrow \left( x+y \right)\left( x-y \right)=0 \\
\end{align}$
Equating each of the two brackets as 0 we get,
$\begin{align}
& x-y=0 \\
& \Rightarrow x=y \\
& x+y=0 \\
& \Rightarrow x=-y \\
\end{align}$
From the above, option (b) is also correct.
So, the correct answer is “Option d”, All of the above.
Note: The mistake that could be possible in the above solution is the calculation mistake and incorrect writing of the trigonometric identities and ratios so make sure you will pay attention while solving the given value of x and y in the above problem and also make sure you have properly written the trigonometric identities and ratios.
Complete step by step answer:
In the above problem, we have given the value of $x\And y$ as follows:
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \sec \left( {{\sin }^{-1}}a \right) \right) \right) \right) \right)$;
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( {{\cos }^{-1}}a \right) \right) \right) \right) \right)$
First of all, we are going to solve x by assuming ${{\sin }^{-1}}a=\theta $ then taking sine on both the sides we get,
$\sin \left( {{\sin }^{-1}}a \right)=\sin \theta $
We know that multiplying any term with its inverse will give us 1 so the evaluation of $\sin \left( {{\sin }^{-1}} \right)=1$ and using this relation in the above we get,
$a=\sin \theta $
Now, putting ${{\sin }^{-1}}a=\theta $ in x we get,
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \sec \theta \right) \right) \right) \right)$
We are going to find the value of $\sec \theta $ using $\sin \theta $ which we have calculated above in the following manner:
$a=\sin \theta $
We know that:
$\sec \theta =\dfrac{1}{\cos \theta }$
And we know the trigonometric identity which states that:
$\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
\end{align}$
Taking square root on both the sides of the above equation we get,
$\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Substituting the value of $\sin \theta =a$ in the above equation we get,
$\cos \theta =\sqrt{1-{{a}^{2}}}$
Using the above relation in $\sec \theta =\dfrac{1}{\cos \theta }$ we get,
$\sec \theta =\dfrac{1}{\sqrt{1-{{a}^{2}}}}$
Substituting the above value in new $x$ we get,
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right) \right) \right) \right)$
Now, we are going to assume ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ and then taking $\cot $ on both the sides of this equation we get,
$\begin{align}
& \cot {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\cot \alpha \\
& \Rightarrow \dfrac{1}{\sqrt{1-{{a}^{2}}}}=\cot \alpha \\
\end{align}$
Substituting ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ in new $x$ we get,
$x=\csc \left( {{\tan }^{-1}}\left( \cos \left( \alpha \right) \right) \right)$
Now, we are going to find the value of $\cos \alpha $ from $\cot \alpha $ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\cot \alpha $
We know that $\cot \alpha $ is the ratio of base (B) to perpendicular (P) the of the right angled triangle so equating the above equation to $\dfrac{B}{P}$ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\cot \alpha =\dfrac{B}{P}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “B” as 1 and “P” as $\sqrt{1-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{\left( \sqrt{1-{{a}^{2}}} \right)}^{2}}+{{1}^{2}}} \\
& \Rightarrow H=\sqrt{1-{{a}^{2}}+1} \\
& \Rightarrow H=\sqrt{2-{{a}^{2}}} \\
\end{align}$
Now, we know that $\cos \alpha =\dfrac{B}{H}$ so substituting the values of “B and H” in this cosine expression and we get,
$\cos \alpha =\dfrac{1}{\sqrt{2-{{a}^{2}}}}$
Substituting the above value of $\cos \alpha $ in new $x$ and we get,
$x=\csc \left( {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right) \right)$
Now, we are going to assume ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ and then taking $\tan $ on both the sides of this equation we get,
$\begin{align}
& \tan {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta \\
& \Rightarrow \left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta \\
\end{align}$
Substituting ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ in new $x$ we get,
$x=\csc \left( \beta \right)$
Now, we are going to find the value of $\csc \beta $ from $\tan \beta $ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta $
We know that $\tan \beta $ is the ratio of perpendicular (P) to the base (B) of the right angled triangle so equating the above equation to $\dfrac{P}{B}$ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\tan \beta =\dfrac{P}{B}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “P” as 1 and “B” as $\sqrt{2-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{1}^{2}}+{{\left( \sqrt{2-{{a}^{2}}} \right)}^{2}}} \\
& \Rightarrow H=\sqrt{1+2-{{a}^{2}}} \\
& \Rightarrow H=\sqrt{3-{{a}^{2}}} \\
\end{align}$
Now, we know that $\csc \beta =\dfrac{H}{P}$ so substituting the values of “H and P” in this cosecant expression and we get,
$\csc \beta =\sqrt{3-{{a}^{2}}}$
Using the above relation in new $x$ we get,
\[x=\sqrt{3-{{a}^{2}}}\]
Taking square on both the sides of the above equation we get,
\[\begin{align}
& {{x}^{2}}={{\left( \sqrt{3-{{a}^{2}}} \right)}^{2}} \\
& \Rightarrow {{x}^{2}}=3-{{a}^{2}} \\
\end{align}\]
Adding ${{a}^{2}}$ on both the sides of the above equation we get,
${{x}^{2}}+{{a}^{2}}=3$ …….. (1)
From the above, option (a) is correct.
Similarly, we are going to solve y in the above problem as follows:
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( {{\cos }^{-1}}a \right) \right) \right) \right) \right)$
First of all, we are going to solve x by assuming ${{\cos }^{-1}}a=\theta $ then taking cosine on both the sides we get,
$\cos \left( {{\cos }^{-1}}a \right)=\theta $
We know that multiplying any term with its inverse will give us 1 so the evaluation of $\cos \left( {{\cos }^{-1}} \right)=1$ and using this relation in the above we get,
$a=\cos \theta $
Now, putting ${{\cos }^{-1}}a=\theta $ in y we get,
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \csc \left( \theta \right) \right) \right) \right) \right)$
We are going to find the value of $\csc \theta $ using $\cos \theta $ which we have calculated above in the following manner:
$a=\cos \theta $
We know that:
$\csc \theta =\dfrac{1}{\sin \theta }$
And we know the trigonometric identity which states that:
$\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& \Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \\
\end{align}$
Taking square root on both the sides of the above equation we get,
$\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$
Substituting the value of $\cos \theta =a$ in the above equation we get,
$\sin \theta =\sqrt{1-{{a}^{2}}}$
Using the above relation in $\csc \theta =\dfrac{1}{\sin \theta }$ we get,
$\csc \theta =\dfrac{1}{\sqrt{1-{{a}^{2}}}}$
Substituting the above value in new y we get,
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right) \right) \right) \right)$
Now, we are going to assume ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ and then taking $\tan $ on both the sides of this equation we get,
$\begin{align}
& \tan {{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\tan \alpha \\
& \Rightarrow \dfrac{1}{\sqrt{1-{{a}^{2}}}}=\tan \alpha \\
\end{align}$
Substituting ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{1-{{a}^{2}}}} \right)=\alpha $ in new y we get,
$y=\sec \left( {{\cot }^{-1}}\left( \sin \left( \alpha \right) \right) \right)$
Now, we are going to find the value of $\sin \alpha $ from $\tan \alpha $ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\tan \alpha $
We know that $\tan \alpha $ is the ratio of perpendicular (P) to base (B) of the right angled triangle so equating the above equation to $\dfrac{P}{B}$ we get,
$\dfrac{1}{\sqrt{1-{{a}^{2}}}}=\tan \alpha =\dfrac{P}{B}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “P” as 1 and “B” as $\sqrt{1-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{1}^{2}}+{{\left( \sqrt{1-{{a}^{2}}} \right)}^{2}}} \\
& \Rightarrow H=\sqrt{1+1-{{a}^{2}}} \\
& \Rightarrow H=\sqrt{2-{{a}^{2}}} \\
\end{align}$
Now, we know that $\sin \alpha =\dfrac{P}{H}$ so substituting the values of “P and H” in this cosine expression and we get,
$\sin \alpha =\dfrac{1}{\sqrt{2-{{a}^{2}}}}$
Substituting the above value of $\sin \alpha $ in new y and we get,
$y=\sec \left( {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right) \right)$
Now, we are going to assume ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ and then taking $\cot $ on both the sides of this equation we get,
$\begin{align}
& \cot {{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta \\
& \Rightarrow \left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta \\
\end{align}$
Substituting ${{\cot }^{-1}}\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\beta $ in new $y$ we get,
$y=\sec \left( \beta \right)$
Now, we are going to find the value of $\sec \beta $ from $\cot \beta $ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta $
We know that $\cot \beta $ is the ratio of base (B) to the perpendicular (P) of the right angled triangle so equating the above equation to $\dfrac{B}{P}$ we get,
$\left( \dfrac{1}{\sqrt{2-{{a}^{2}}}} \right)=\cot \beta =\dfrac{B}{P}$
Also, hypotenuse is equal to the square root of the addition of the square of perpendicular and the square of the base so hypotenuse (H) is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting “B” as 1 and “P” as $\sqrt{2-{{a}^{2}}}$ in the above equation we get,
$\begin{align}
& H=\sqrt{{{\left( \sqrt{2-{{a}^{2}}} \right)}^{2}}+{{1}^{2}}} \\
& \Rightarrow H=\sqrt{2-{{a}^{2}}+1} \\
& \Rightarrow H=\sqrt{3-{{a}^{2}}} \\
\end{align}$
Now, we know that $\sec \beta =\dfrac{H}{B}$ so substituting the values of “H and B” in this secant expression and we get,
$\sec \beta =\sqrt{3-{{a}^{2}}}$
Using the above relation in new $y$ we get,
\[y=\sqrt{3-{{a}^{2}}}\]
Taking square on both the sides of the above equation we get,
\[\begin{align}
& {{y}^{2}}={{\left( \sqrt{3-{{a}^{2}}} \right)}^{2}} \\
& \Rightarrow {{y}^{2}}=3-{{a}^{2}} \\
\end{align}\]
Adding ${{a}^{2}}$ on both the sides of the above equation we get,
\[{{y}^{2}}+{{a}^{2}}=3\]………… (2)
From the above option (c) is correct.
From eq. (1) and eq. (2) we get,
\[\begin{align}
& {{x}^{2}}+{{a}^{2}}=3 \\
& {{y}^{2}}+{{a}^{2}}=3 \\
\end{align}\]
Rearranging the above two equations we get,
\[\begin{align}
& {{x}^{2}}=3-{{a}^{2}} \\
& {{y}^{2}}=3-{{a}^{2}} \\
\end{align}\]
From the above, ${{x}^{2}}={{y}^{2}}$ so subtracting ${{y}^{2}}$ on both the sides of this equation we get,
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=0 \\
& \Rightarrow \left( x+y \right)\left( x-y \right)=0 \\
\end{align}$
Equating each of the two brackets as 0 we get,
$\begin{align}
& x-y=0 \\
& \Rightarrow x=y \\
& x+y=0 \\
& \Rightarrow x=-y \\
\end{align}$
From the above, option (b) is also correct.
So, the correct answer is “Option d”, All of the above.
Note: The mistake that could be possible in the above solution is the calculation mistake and incorrect writing of the trigonometric identities and ratios so make sure you will pay attention while solving the given value of x and y in the above problem and also make sure you have properly written the trigonometric identities and ratios.
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