Prove that:
\[\dfrac{{\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta }}{{\tan \theta }} - 1 = - {\sin ^2}\theta \].

Hint: While approaching such kinds of questions first of all we have to see what are provided with and what we have to prove. Here we are provided with the left hand side that is \[\dfrac{{\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta }}{{\tan \theta }} - 1\]and we have to prove that it is equal to the right hand side that is \[ - {\sin ^2}\theta \]. Hence for start of the solution we need to do the simplification of the term in the numerator of the left hand side that is as a starting part \[\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta \]by using the trigonometry identity of \[\sin \left( {{{90}^\circ } - \theta } \right) = \cos \theta \] later further solving the left hand side to get the desired result let us see the whole application of this in the complete step by step solution.

Complete Step by Step Solution:
Here we are asked in the above question to prove that \[\dfrac{{\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta }}{{\tan \theta }} - 1 = - {\sin ^2}\theta \]
For this taking the left hand side of the \[\dfrac{{\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta }}{{\tan \theta }} - 1 = - {\sin ^2}\theta \] is \[\dfrac{{\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta }}{{\tan \theta }} - 1\]
So firstly simplifying the numerator in the left hand side that is \[\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta \]
Now using the Basic trigonometric identity of \[\sin \left( {{{90}^\circ } - \theta } \right) = \cos \theta \]in the numerator of left hand side we get –
\[\dfrac{{\cos \theta \sin \theta }}{{\tan \theta }} - 1\]
Now breaking the \[\tan \theta \]into it’s components that is \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \]and putting it in the above equation we get-
\[
  \dfrac{{\cos \theta \sin \theta }}{{\dfrac{{\sin \theta }}{{\cos \theta }}}} - 1 \\
  \dfrac{{{{\cos }^2}\theta \sin \theta }}{{\sin \theta }} - 1 \\
 \]
Further simplifying we get –
\[
  \dfrac{{{{\cos }^2}\theta }}{{}} \times \dfrac{{\sin \theta }}{{\sin \theta }} - 1 \\
  {\cos ^2}\theta - 1 \\
 \]
Now here we would use the identity of trigonometry that is –
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Further this identity can be derived into of form –
\[{\cos ^2}\theta - 1 = - {\sin ^2}\theta \]
So using above derivation of identity in the resulting equation of left hand side that has come out to be –
 \[{\cos ^2}\theta - 1\]
Which is equal to the \[ - {\sin ^2}\theta \]which is also equal to the term given in the right hand side that is \[ - {\sin ^2}\theta \]
So on a conclusion left hand side is same to the right hand side
Hence proved
\[
  \dfrac{{\sin \left( {{{90}^\circ } - \theta } \right)\sin \theta }}{{\tan \theta }} - 1 = - {\sin ^2}\theta \\
    \\
    \\
 \]

Note :
While solving such kind of the questions one should keep in mind the basic trigonometric identities like \[\sin \left( {{{90}^\circ } - \theta } \right) = \cos \theta \]and \[\cos \left( {{{90}^\circ } - \theta } \right) = \sin \theta \]also \[\tan \left( {{{90}^\circ } - \theta } \right) = \cot \theta \]
And the other identities of the usage like\[{\sin ^2}\theta + {\cos ^2}\theta = 1\],\[1 + {\cot ^2}\theta = \cos e{c^2}\theta \]and \[1 + {\tan ^2}\theta = se{c^2}\theta \]remembering these kind of equations help to solve the questions easily and rapidly.