${\text{si}}{{\text{n}}^2}5^\circ + {\text{si}}{{\text{n}}^2}10^\circ + {\text{si}}{{\text{n}}^2}15^\circ .... + {\text{si}}{{\text{n}}^2}85^\circ + {\text{si}}{{\text{n}}^2}90^\circ = 9\dfrac{1}{2}$

Hint: In order to prove this we use the trigonometric identities ${\text{sin}}\left( {90 - \theta } \right) = {\text{cos}}\theta $and ${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$ to rearrange the terms and solve.

Complete step-by-step answer:
We consider the LHS and convert into RHS
We have, the LHS is
${\text{si}}{{\text{n}}^2}5^\circ + {\text{si}}{{\text{n}}^2}10^\circ + {\text{si}}{{\text{n}}^2}15^\circ .... + {\text{si}}{{\text{n}}^2}85^\circ + {\text{si}}{{\text{n}}^2}90^\circ $
We use the LHS and try to express it as RHS to prove the given.

We know that from the identity of sine and cos function,
${\text{sin}}\left( {90 - \theta } \right) = {\text{cos}}\theta $.
Squaring on both sides, we get
So ${\text{si}}{{\text{n}}^2}\left( {90 - \theta } \right) = {\text{co}}{{\text{s}}^2}\theta $

Now we use the above identity to express the LHS of the given as follows:
$ \Rightarrow {\text{si}}{{\text{n}}^2}5^\circ + {\text{si}}{{\text{n}}^2}10^\circ + {\text{si}}{{\text{n}}^2}15^\circ ... + {\text{si}}{{\text{n}}^2}45^\circ ....{\text{si}}{{\text{n}}^2}\left( {90 - 15} \right)^\circ + {\text{si}}{{\text{n}}^2}\left( {90 - 10} \right)^\circ + {\text{si}}{{\text{n}}^2}\left( {90 - 5} \right)^\circ + {\text{si}}{{\text{n}}^2}90^\circ $$ \Rightarrow {\text{si}}{{\text{n}}^2}5^\circ + {\text{si}}{{\text{n}}^2}10^\circ + {\text{si}}{{\text{n}}^2}15^\circ ... + {\text{si}}{{\text{n}}^2}45^\circ ....{\text{co}}{{\text{s}}^2}\left( {15} \right)^\circ + {\text{co}}{{\text{s}}^2}\left( {10} \right)^\circ + {\text{co}}{{\text{s}}^2}\left( 5 \right)^\circ + {\text{si}}{{\text{n}}^2}90^\circ $
Each term after 45° of angle in sin function, is expressed as a difference of 90° and its complementary angle as shown.

Also we know the identity, ${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$
So we rearrange the terms in the given equation we get 8 cases of ${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$
$ \Rightarrow {\text{si}}{{\text{n}}^2}5^\circ + {\text{co}}{{\text{s}}^2}5^\circ + {\text{si}}{{\text{n}}^2}15^\circ + {\text{co}}{{\text{s}}^2}15^\circ + ...........{\text{si}}{{\text{n}}^2}45^\circ + {\text{si}}{{\text{n}}^2}90^\circ $
$
   \Rightarrow 8{\text{ + si}}{{\text{n}}^2}45^\circ + {\text{si}}{{\text{n}}^2}90^\circ \\
   \Rightarrow 8 + \dfrac{1}{2} + 1 \\
   \Rightarrow 9\dfrac{1}{2} \\
$
Hence LHS = RHS, hence proved.

Note: In order to solve problems of this type the key is to pick the appropriate identity such that when applied makes the equation the way we want it. Having adequate knowledge in trigonometric identities and trigonometric table of sine, is required.