Question

Prove that ${\tan }^{2}a-{\tan }^{2}b=\frac{[{\cos }^{2}b-{\cos }^{2}a]}{{\cos }^{2}a{\cos }^{2}b}=\frac{({\sin }^{2}a-{\sin }^{2}b)}{({\cos }^{2}a.{\cos }^{2}b)}$

Answer 1

Determine the proof of the given expression ${\tan }^{2}a-{\tan }^{2}b=\frac{[{\cos }^{2}b-{\cos }^{2}a]}{{\cos }^{2}a{\cos }^{2}b}=\frac{({\sin }^{2}a-{\sin }^{2}b)}{({\cos }^{2}a.{\cos }^{2}b)}$

Solve the L.H.S part:

$$\begin{gathered} {\tan }^{2}a–{\tan }^{2}b=\left(\frac{{\sin }^{2}a}{{\cos }^{2}a}\right)–\left(\frac{{\sin }^{2}b}{{\cos }^{2}b}\right) \\ \Rightarrow =\frac{({\cos }^{2}b{\sin }^{2}a–{\cos }^{2}a{\sin }^{2}b)}{{\cos }^{2}a{\cos }^{2}b} \\ \Rightarrow =\frac{[{\cos }^{2}b(1–{\cos }^{2}a)–{\cos }^{2}a(1–{\cos }^{2}b\left)\right]}{{\cos }^{2}a{\cos }^{2}b}\mathbf{\because }\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{}\mathit{a}=1 \\ \Rightarrow =\frac{[{\cos }^{2}b–{\cos }^{2}b{\cos }^{2}a–{\cos }^{2}a+{\cos }^{2}a{\cos }^{2}b]}{{\cos }^{2}a{\cos }^{2}b} \\ \Rightarrow =\frac{[{\cos }^{2}b–{\cos }^{2}a]}{{\cos }^{2}a{\cos }^{2}b} \\ \Rightarrow =\frac{[1–{\sin }^{2}b–1+{\sin }^{2}a]}{{\cos }^{2}a{\cos }^{2}b} \\ \Rightarrow =\frac{({\sin }^{2}a–{\sin }^{2}b)}{{\cos }^{2}a{\cos }^{2}b} \end{gathered}$$

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