Question

Prove that $\frac{\tan A}{(1-cotA)}+\frac{cotA}{(1-\tan A)}=1+secA.\cos ecA$.

Answer 1

To prove: $\frac{\tan A}{(1-cotA)}+\frac{cotA}{(1-\tan A)}=1+secA.\cos ecA$

Consider L.H.S :

$$\begin{gathered} \frac{\tan A}{(1-\cot A)}+\frac{\cot A}{(1-\tan A)}=\frac{\tan A}{(1-\frac{1}{\mathrm{ta}nA})}+\frac{\frac{1}{\tan A}}{(1-\tan A)} \\ \Rightarrow =\frac{{\tan }^{2}A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)} \\ \Rightarrow =\frac{1-{\tan }^{3}A}{\tan A(1-\tan A)} \\ \Rightarrow =\frac{(1-\tan A)(1+\tan A+{\tan }^{2}A)}{\tan A(1-\tan A)}\because {\mathit{a}}^{\mathbf{3}}\mathbf{-}{\mathit{b}}^{\mathbf{3}}=(a-b)(a^2+ab+b^2) \\ \Rightarrow =\frac{{\sec }^{2}A+\tan A}{\tan A}\because \mathbf{1}\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{A}={\sec }^{2}A \\ \Rightarrow =1+\frac{{\sec }^{2}A}{\tan A} \\ \Rightarrow =1+\frac{1}{\cos A\sin A} \\ \Rightarrow =1+\sec AcosecA \\ =R.H.S \end{gathered}$$

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