Answer the 7th question

$$\begin{gathered} \mathrm{LHS} = {\tan }^2\mathrm{A} - {\tan }^2\mathrm{B} \\ =\frac{{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}} - \frac{{\sin }^2\mathrm{B}}{{\cos }^2\mathrm{B}} \\ =\frac{{\sin }^2\mathrm{A} . {\cos }^2\mathrm{B} - {\cos }^2\mathrm{A} . {\sin }^2\mathrm{B}}{{\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}} \\ =\frac{\left(1-{\cos }^2\mathrm{A}\right) . {\cos }^2\mathrm{B} - {\cos }^2\mathrm{A}\left(1 - {\cos }^2\mathrm{B}\right)}{{\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}} \\ =\frac{{\cos }^2\mathrm{B}- {\cos }^2\mathrm{A} . {\cos }^2\mathrm{B} - {\cos }^2\mathrm{A} + {\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}}{{\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}} \\ =\frac{{\cos }^2\mathrm{B} - {\cos }^2\mathrm{A}}{{\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}} \\ =\frac{1 - {\sin }^2\mathrm{B} - 1 + {\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}} \\ =\frac{{\sin }^2\mathrm{A} - {\sin }^2\mathrm{B}}{{\cos }^2\mathrm{A} . {\cos }^2\mathrm{B}} \\ =\mathrm{RHS} \end{gathered}$$