How do you verify the given result $ {{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta {{\sin }^{2}}\theta $ ?

Hint: We start solving the problem by considering the L.H.S (Left Hand Side) of the given result. We then make use of the fact $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ to proceed through the problem. We then make the necessary calculations and make use of the result $ {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ to proceed further through the problem. We then make the necessary calculations and make use of the fact $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ to complete the required proof.

Complete step by step answer:
According to the problem, we are asked to verify the given result $ {{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta {{\sin }^{2}}\theta $ .
Let us first consider L.H.S (Left Hand Side) of the given result and prove it equal to the R.H.S to complete the verification.
So, let us consider L.H.S (Left Hand Side) which is $ {{\tan }^{2}}\theta -{{\sin }^{2}}\theta $ ---(1).
We know that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ . Let us use this result in equation (1).
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta \].
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }\].
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta \left( 1-{{\cos }^{2}}\theta \right)}{{{\cos }^{2}}\theta }\] ---(2).
We know that $ {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ . Let us use this result in equation (2).
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta \left( {{\sin }^{2}}\theta \right)}{{{\cos }^{2}}\theta }\].
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\left( {{\sin }^{2}}\theta \right)\].
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}\left( {{\sin }^{2}}\theta \right)\] ---(3).
We know that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ . Let us use this result in equation (3).
\[\Rightarrow {{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta {{\sin }^{2}}\theta \] ---(4).
From equation (4), we can see that we have proved that the given L.H.S (Left Hand Side) is equal to the R.H.S (Right Hand Side) given in the problem.
$ \therefore $ We have proved the result $ {{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta {{\sin }^{2}}\theta $ .

Note:
 We can also verify the given result by substituting the angles in it and checking the values on both sides of the given equation. We should not make calculation mistakes while solving this problem. Whenever we get this type of problems, we first consider L.H.S (Left Hand Side) of the given result and then try to prove it equal to the R.H.S (Right Hand Side). Similarly, we can expect problems to prove the result \[\text{cose}{{\text{c}}^{2}}\theta +\text{se}{{\text{c}}^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta \text{se}{{\text{c}}^{2}}\theta \].