Show that \[{{\sin }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{10}^{\circ }}+{{\sin }^{2}}{{15}^{\circ }}+..............+{{\sin }^{2}}{{90}^{\circ }}=9\dfrac{1}{2}\]

Hint: We will first simplify all the terms in the expression mentioned in the question in multiples of 90 plus some angle and then with the help of cofunction identities we will convert these terms into simple function with standard angles and then we will use \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] a few times and after applying this we will solve this question.
Complete step-by-step answer:
We will begin with the left hand side of the expression mentioned in the question,\[\Rightarrow {{\sin }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{10}^{\circ }}+{{\sin }^{2}}{{15}^{\circ }}+..............+{{\sin }^{2}}{{90}^{\circ }}........(1)\]
Now writing some more terms in the series in equation (1) we get,
\[\Rightarrow {{\sin }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{10}^{\circ }}+{{\sin }^{2}}{{15}^{\circ }}+........+{{\sin }^{2}}{{45}^{\circ }}......+{{\sin }^{2}}{{75}^{\circ }}+{{\sin }^{2}}{{80}^{\circ }}+{{\sin }^{2}}85+{{\sin }^{2}}{{90}^{\circ }}.......(2)\]
Now converting some of the angles in terms of 90 in equation (2) we get,
\[\Rightarrow {{\sin }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{10}^{\circ }}+{{\sin }^{2}}{{15}^{\circ }}+.......+{{\sin }^{2}}({{90}^{\circ }}-{{15}^{\circ }})+{{\sin }^{2}}({{90}^{\circ }}-{{10}^{\circ }})+{{\sin }^{2}}({{90}^{\circ }}-{{85}^{\circ }})+{{\sin }^{2}}{{90}^{\circ }}.......(3)\]
We know that \[\sin (90-\theta )=\cos \theta \] and when we square both sides we get \[{{\sin }^{2}}(90-\theta )={{\cos }^{2}}\theta \] and hence applying this in equation (3) we get,
\[\Rightarrow {{\sin }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{10}^{\circ }}+{{\sin }^{2}}{{15}^{\circ }}+.......+{{\cos }^{2}}{{15}^{\circ }}+{{\cos }^{2}}{{10}^{\circ }}+{{\cos }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{90}^{\circ }}.......(4)\]
So now in the given series in equation (4) on rearranging terms we get 8 cases where \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}{{5}^{\circ }}+{{\cos }^{2}}{{5}^{\circ }}+{{\sin }^{2}}{{10}^{\circ }}+{{\cos }^{2}}{{10}^{\circ }}+{{\sin }^{2}}{{15}^{\circ }}+{{\cos }^{2}}{{15}^{\circ }}+.......{{\sin }^{2}}{{45}^{\circ }}+{{\sin }^{2}}{{90}^{\circ }} \\
 & \Rightarrow 1+1+1+1+1+1+1+1+{{\sin }^{2}}{{45}^{\circ }}+{{\sin }^{2}}{{90}^{\circ }}............(5) \\
\end{align}\]
Now substituting all the values of the standard angles in equation (5) we get,
\[\Rightarrow 8+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{1}^{2}}=9+\dfrac{1}{2}=\dfrac{19}{2}=9\dfrac{1}{2}.........(6)\]
Hence from equation (6) we can see that the value is equal to the right hand side of the expression mentioned in the question.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identities as we can write cos in place of sin and sin in place of cos in equation (3).