Question

Prove That $ta{n}^{2}A-si{n}^{2}A=ta{n}^{2}A\times si{n}^{2}A$

Answer 1

Determine the proof of the given expression that is $ta{n}^{2}A-si{n}^{2}A=ta{n}^{2}A\times si{n}^{2}A$.

Solve the L.H.S part:

$$\begin{gathered} ta{n}^{2}A-si{n}^{2}A=\frac{{\sin }^{2}A}{{\cos }^{2}A}-si{n}^{2}A \\ \Rightarrow =\frac{{\sin }^{2}A-{\sin }^{2}A.{\cos }^{2}A}{{\cos }^{2}A} \\ \Rightarrow =\frac{{\sin }^{2}A(1-{\cos }^{2}A)}{{\cos }^{2}A}\mathbf{\because }\mathbf{}\mathbf{}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{}\mathit{A}=1 \\ \Rightarrow =\frac{{\sin }^{2}A.{\sin }^{2}A}{{\cos }^{2}A} \\ \Rightarrow ={\sin }^{2}A\times \frac{{\sin }^{2}A}{{\cos }^{2}A} \\ \Rightarrow ={\sin }^{2}A.{\tan }^{2}A \end{gathered}$$

Hence, the L.H.S = R.H.S.