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If we use the limit formula we have that as $xto pi/2$ from the left we have a positive slope and as it approaches from the right we have a negative slope.
So, the right hand limit and left hand limit are different. How is it differentiable at the point then ?
We use the same logic and say $|x|$ is not differentiable at $x=0$

To respond to your comments:

The limit values may be almost same but their angles differ right ? Like for x=π/2 approaching from the left is gives us angle close to 0 while x=π/2 approaching from the right gives us an angle close to 180 degrees.

It's not correct that "the limit values are almost the same". The left-hand limit is exactly $0$, no more, no less. The right-hand limit is also exactly $0$, no more, no less. Since the left-hand limit and the right-hand limit are equal (exactly equal), the limit exists.

It's true that "the angles differ", but it also doesn't matter. The definition of a derivative doesn't involve or mention angles at all.

This is a logical question. However, you can show the limit is equivalent for both the LHS and RHS by

$$lim_{x to frac{pi}{2}^+}sin’x = cosbigg(frac{pi}{2}^+bigg) = 0$$

$$lim_{x to frac{pi}{2}^-}sin’x = cosbigg(frac{pi}{2}^-bigg) = 0$$

Let’s put this example aside for a second and look at $y = vert xvert$ and $y = x^2$, which are similar to your example in that the slopes seem to differ at $x > 0$ and $x < 0$. Here, the derivative doesn’t exist for $y = vert xvert$ since the RHS and LHS limits are not equal, but the derivative does exist for $y = x^2$, so what is the difference?

You can observe the reason visually without calculating. Check https://www.desmos.com/calculator/czqmiiuxrr (the graph the functions). Then, try to zoom in on $x = 0$. Notice that $y = x^2$ seems to flatten out the more you zoom in. However, no matter how much you zoom in on $y = vert xvert$, it always retains its V-shape.

$sin x$ behaves like $y = x^2$ in this regard. It flattens out if you zoom in on $x = frac{pi}{2}$, so the slope of the tangent at that point is $0$. You can check https://www.desmos.com/calculator/l5lv0sypss (the graph of $sin x$) to see what I mean.

From the left it approaches to $0$, and from the right it approaches to $0$ as well.

Compare this to $|x|$, from the left it approaches to $-1$, and from the right it goes to $+1$.

Do you see the difference?

We have that

$$lim_{xto frac {pi}2} frac{sin x-sin frac {pi}2}{x-frac {pi}2}
=lim_{yto 0} frac{sin left( frac {pi}2+yright)-1}{y}
=lim_{yto 0} frac{cos y-1}{y}=lim_{yto 0} left(ycdotfrac{cos y-1}{y^2}right)=0cdot left(-frac12right)=0$$

The limit is 0 each case. In one case the limit is approached through positive values, and in the other the limit is approached through negative values; but in each case the limit itself is the same: zero.

It looks you are considering the signs of the numerator and denominator in:
$$lim_{xto frac{pi}{2}^-} frac{sin(x)-sin{frac{pi}{2}}}{x-frac{pi}{2}} text{and} lim_{xto frac{pi}{2}^+} frac{sin(x)-sin{frac{pi}{2}}}{x-frac{pi}{2}}.$$
In the LHS limit, both numerator and denominator are negative, while in the RHS limit, the numerator is negative and denominator is positive, which seems a contradiction to you. ("So, the right hand limit and left hand limit are different.") The thing is, as others pointed, both LHS and RHS limits are eventually equal to zero.

As an easier example, consider the derivative of $f(x)=x^2$ at $x=0$. By your reasoning the function must not have a derivative, while it does have it, because:
$$lim_{xto 0^-} frac{x^2-0}{x-0}=0 text{and} lim_{xto 0^+} frac{x^2-0}{x-0}=0.$$