Question

Prove that
1-tan×tan÷cot×cot-1=tan×tan

Answer 1

hope its helps u bro

Answer 2

$\; \star \; {\underline{\boxed{\pmb{\orange{\frak{ To Prove\; :- }}}}}}$

$\sf \longmapsto{ \dfrac{ 1 - \tan {}^{2} }{ \cot {}^{2} - 1 } = \tan {}^{2} }$

$\begin{gathered}\begin{gathered} \\ \\ \end{gathered} \end{gathered}$

$\; \star \; {\underline{\boxed{\pmb{\purple{\frak{ \; SolutioN \; :- }}}}}}$

$\begin{gathered}\begin{gathered} \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{1 - \tan {}^{2} }{ \cot {}^{2} - 1 } } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{1 - \dfrac{ \sin {}^{2}}{ \cos {}^{2} } }{ \dfrac{\cos{}^{2}}{ \sin {}^{2} } - 1 } } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{ \dfrac{ \cos { }^{2} - \sin {}^{2}}{ \cos {}^{2} } }{ \dfrac{\cos{}^{2} - \sin {}^{2} } { \sin {}^{2} } } } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{ \dfrac{ \cancel{\cos { }^{2} - \sin {}^{2}}}{ \cos {}^{2} } }{ \dfrac{ \cancel{\cos{}^{2} - \sin {}^{2}} } { \sin {}^{2} } } } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{ \dfrac{ 1}{ \cos {}^{2} } }{ \dfrac{ 1 } { \sin {}^{2} } } } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{ { 1}{ } }{ \cos {}^{2} } \times \sin {}^{2} } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { \dfrac{ { \sin {}^{2} }{ } }{ \cos {}^{2} } } \\ \\ \end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \; \longmapsto \; \sf { { \tan {}^{2} } } \\ \\ \end{gathered}\end{gathered}$

Hence Proved.

$\begin{gathered}\begin{gathered} \\ {\underline{\rule{300pt}{9pt}}} \end{gathered} \end{gathered}$