Maximum value of sinx-cosx is equal to
\[\begin{align}
  & A.\sqrt{2} \\
 & B.1 \\
 & C.0 \\
 & D.\text{None of these} \\
\end{align}\]

Hint: In this question, we need to find the maximum value of sinx-cosx. Suppose $f\left( x \right)=\sin x-\cos x$ for this, we will use a second derivative test. We will first find the derivative of a given function i.e. f'(x). Then we will put f'(x) = 0 and find the value of x. Then we will again find the derivative of the function i.e. f''(x). Using value of x found before in f''(x). We will check if the point gives maximum value or minimum value. If f''(x) < 0 then x gives maximum value and if f''(x) > 0 then x gives minimum value. Using the value of x which gives maximum value we will find maximum f(x). We will use following properties,
$\begin{align}
  & \left( i \right)\dfrac{d}{dx}\sin x=\cos x \\
 & \left( ii \right)\dfrac{d}{dx}\cos x=-\sin x \\
 & \left( iii \right)\tan x=\dfrac{\sin x}{\cos x} \\
 & \left( iv \right)\sin \left( \pi -\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4}\text{ and }\cos \left( \pi -\dfrac{\pi }{4} \right)=-\cos \dfrac{\pi }{4} \\
\end{align}$

Complete step by step answer:
Here we are given the function as sinx-cosx. Let us suppose it to be equal to f(x).
We get $f'\left( x \right)=\dfrac{d}{dx}\left( \sin x-\cos x \right)$.
Now let us find the derivative of a given function.
Differentiating both sides w.r.t x, we get $f'\left( x \right)=\dfrac{d}{dx}\left( \sin x-\cos x \right)$.
We know that $\dfrac{d}{dx}\sin x=\cos x\text{ and }\dfrac{d}{dx}\cos x=-\sin x$ so we get, $f'\left( x \right)=\cos x-\left( -\sin x \right)\Rightarrow f'\left( x \right)=\cos x+\sin x\cdots \cdots \cdots \left( 2 \right)$.
Now again let us find the derivative.
Differentiating w.r.t x we get $f''\left( x \right)=\dfrac{d}{dx}\left( \cos x+\sin x \right)$.
Again using $\dfrac{d}{dx}\sin x=\cos x\text{ and }\dfrac{d}{dx}\cos x=-\sin x$ we get, $f''\left( x \right)=-\sin x+\cos x\cdots \cdots \cdots \left( 3 \right)$.
Now let us equate equation (2) to zero to find the value of x, we get $f'\left( x \right)=0\Rightarrow \cos x+\sin x=0\Rightarrow \cos x=-\sin x$.
Dividing by cos x on both sides we get, $1=\dfrac{-\sin x}{\cos x}$.
Using the property of $\tan \theta $ i.e. $\tan x=\dfrac{\sin x}{\cos x}$ we get, $\tan x=-1$.
We know that $\tan \theta $ is negative in the second quadrant and $\tan \dfrac{\pi }{4}=1$. So in the second quadrant we will have $\tan \left( \pi -\dfrac{\pi }{4} \right)=-\tan \dfrac{\pi }{4}=-1$.
Hence $\tan \dfrac{3\pi }{4}=-1$.
Therefore, $x=\dfrac{3\pi }{4}$.
Let us check if it gives maximum value or not.
Putting $x=\dfrac{3\pi }{4}$ in (3) we get, $f''\left( \dfrac{3\pi }{4} \right)=-\sin \dfrac{3\pi }{4}+\cos \dfrac{3\pi }{4}$.
It can be written as $-\sin \left( \pi -\dfrac{\pi }{4} \right)+\cos \left( \pi -\dfrac{\pi }{4} \right)$.
We know $\sin \left( \pi -\theta \right)=\sin \theta \text{ and }\cos \left( \pi -\theta \right)=-\cos \theta $ so we get,
$f''\left( \dfrac{3\pi }{4} \right)=-\sin \dfrac{\pi }{4}-\cos \dfrac{\pi }{4}$.
Since $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}=\cos \dfrac{\pi }{4}$ so we get, $f''\left( \dfrac{3\pi }{4} \right)=-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\Rightarrow f''\left( \dfrac{3\pi }{4} \right)=-\dfrac{2}{\sqrt{2}}\text{ }<\text{ }0$.
So $x=\dfrac{3\pi }{4}$ gives the maximum value of f(x).
Putting $x=\dfrac{3\pi }{4}$ in f(x) we get $f\left( \dfrac{3\pi }{4} \right)=\sin \dfrac{3\pi }{4}-\cos \dfrac{3\pi }{4}$.
As evaluated earlier $\sin \dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}}\text{ and }\cos \dfrac{3\pi }{4}=-\dfrac{1}{\sqrt{2}}$ so we get,
$f\left( \dfrac{3\pi }{4} \right)=\dfrac{1}{\sqrt{2}}-\left( -\dfrac{1}{\sqrt{2}} \right)\Rightarrow \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}$.
Splitting 2 into $\sqrt{2}\times \sqrt{2}$ we get,
$f\left( \dfrac{3\pi }{4} \right)=\dfrac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}=\sqrt{2}$.
Hence the maximum value of sinx-cosx is $\sqrt{2}$.

So, the correct answer is “Option A”.

Note: Students should know the values of $\tan \dfrac{\pi }{4},\sin \dfrac{\pi }{4},\cos \dfrac{\pi }{4}$ from the trigonometric ratio table. Note that there are many more values of $\theta $ for which $\tan \theta =-1$. They can be found by formula $\tan \theta =\tan \alpha \Rightarrow \theta =n\pi \pm \alpha ,n=0,1,2,\ldots \ldots $.