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# If $\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}$ where $x\in [0,\pi ]$ then,a.$x=\dfrac{\pi }{4},y=1$b.$x=\dfrac{\pi }{4},y=0$c.$x=1,y=1$d.$x=\dfrac{3\pi }{4},y=1$

Hint: We will first of all use the fact that the minimum value of $y+\dfrac{1}{y}$ is 2. After this we will apply necessary properties of trigonometric identities such as ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ or any other identities to solve the question completely.

We are given that if $\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}$where$x\in [0,\pi ]$ then we have to find the value of x and y.

To proceed the question, we use the given condition $\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}$ and try to find the value of x and y.

We know that the minimum value of $y+\dfrac{1}{y}$ is 2 that is,
$y+\dfrac{1}{y}\ge 2$
$\Rightarrow \sqrt{y+\dfrac{1}{y}}\ge \sqrt{2}$

Using the obtained expression in the value of $\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}$ we get,
\begin{align} & \Rightarrow \sin x+\cos x=\sqrt{y+\dfrac{1}{y}}\ge \sqrt{2} \\ & \Rightarrow \sin x+\cos x\ge \sqrt{2} \\ \end{align}

Squaring both sides of the above expression,
\begin{align} & \Rightarrow {{\left( \sin x+\cos x \right)}^{2}}\ge 2 \\ & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x\ge 2 \\ \end{align}

We have a trigonometric identity given as,
${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Using this identity in the above obtained expression we get,
$1+2\sin x\cos x\ge 2$

We will again use a trigonometric identity given as,
$2\sin x\cos x=\sin 2x$

Using this identity in the above obtained expression we get,
\begin{align} & 1+\sin 2x\ge 2 \\ & \Rightarrow \sin 2x\ge 1 \\ \end{align}

Now because the maximum value of sin is equal to 1, therefore $\sin 2x\ge 1\Rightarrow \sin 2x=1$
\begin{align} & \Rightarrow \sin 2x=1 \\ & \Rightarrow 2x=\dfrac{\pi }{2} \\ & \Rightarrow x=\dfrac{\pi }{4} \\ \end{align}

Hence, we got $x=\dfrac{\pi }{4}$ thus our answer reduces to two options (a) and (b).

Now observing option (b) we get y=0, which is not possible because this y=0 makes the term $\sqrt{y+\dfrac{1}{y}}$ tends to infinity as we have $\dfrac{1}{y}$tends to infinity as y tends to 0.

Hence, we get our answer as $x=\dfrac{\pi }{4}$ and y=0 which is option (a).

Note: The possibility of error in this question is that when you obtain $x=\dfrac{\pi }{4}$ as result you may go for opting y=0 as the answer, which is wrong because y=0 will make the expression $\sqrt{y+\dfrac{1}{y}}$ tends to infinity as we have $\dfrac{1}{y}$ tends to infinity when y=0.