T^n = sin^n + cos^n. to …

We know,
{tex}{T_n} = {\sin ^n}\theta + {\cos ^n}\theta {/tex}
{tex}\therefore {T_3} = {\sin ^3}\theta + {\cos ^3}\theta {/tex}
{tex}{T_5} = {\sin ^5}\theta + {\cos ^5}\theta {/tex}
{tex}{T_1} = \sin \theta + \cos \theta {/tex}
{tex}{T_7} = {\sin ^7}\theta + {\cos ^7}\theta {/tex}+ cot{tex}\theta{/tex} = m and cosec{tex}\theta{/tex} - cot{tex}\theta{/tex} = n,
LHS {tex} = \frac{{{T_3} - {T_5}}}{{{T_1}}}{/tex}
{tex} = \frac{{{{\sin }^3}\theta + {{\cos }^3}\theta - \left( {{{\sin }^5}\theta + {{\cos }^5}\theta } \right)}}{{\sin \theta + \cos \theta }}{/tex}
{tex} = \frac{{{{\sin }^3}\theta + {{\cos }^3}\theta - {{\sin }^5}\theta - {{\cos }^5}\theta }}{{\sin \theta + \cos \theta }}{/tex}
{tex} = \frac{{{{\sin }^3}\theta - {{\sin }^5}\theta + {{\cos }^3}\theta - {{\cos }^5}\theta }}{{\sin \theta + \cos \theta }}{/tex}
{tex} = \frac{{{{\sin }^3}\theta (1 - {{\sin }^2}\theta ) + {{\cos }^3}\theta (1 - {{\cos }^2}\theta )}}{{\sin \theta + \cos \theta }}{/tex}
{tex} = \frac{{{{\sin }^3}\theta {{\cos }^2}\theta + {{\cos }^3}\theta {{\sin }^2}\theta }}{{\sin \theta + \cos \theta }}{/tex} {tex}\left[ \begin{gathered} \because 1 - {\sin ^2}\theta = {\cos ^2}\theta \hfill \\ 1 - {\cos ^2}\theta = {\sin ^2}\theta \hfill \\ \end{gathered} \right]{/tex}
{tex} = \frac{{{{\sin }^2}\theta {{\cos }^2}\theta (\sin \theta + \cos \theta )}}{{(\sin \theta + \cos \theta )}}{/tex}
{tex} = {\sin ^2}\theta {\cos ^2}\theta {/tex}
 RHS {tex} = \frac{{{T_5} - {T_7}}}{{{T_3}}}{/tex}
{tex} = \frac{{{{\sin }^5}\theta + {{\cos }^5}\theta - ({{\sin }^7}\theta + {{\cos }^7}\theta )}}{{\left( {{{\sin }^3}\theta + {{\cos }^3}\theta } \right)}}{/tex}
{tex} = \frac{{{{\sin }^5}\theta + {{\cos }^5}\theta - {{\sin }^7}\theta - {{\cos }^7}\theta }}{{({{\sin }^3}\theta + {{\cos }^3}\theta )}}{/tex}
{tex} = \frac{{{{\sin }^5}\theta - {{\sin }^7}\theta + {{\cos }^5}\theta - {{\cos }^7}\theta }}{{{{\sin }^3}\theta + {{\cos }^3}\theta }}{/tex}
{tex} = \frac{{{{\sin }^5}\theta \left( {1 - {{\sin }^2}\theta } \right) + {{\cos }^5}\theta \left( {1 - {{\cos }^2}\theta } \right)}}{{{{\sin }^3}\theta + {{\cos }^3}\theta }}{/tex}
{tex} = \frac{{{{\sin }^5}\theta {{\cos }^2}\theta + {{\cos }^5}\theta {{\sin }^2}\theta }}{{{{\sin }^3}\theta + {{\cos }^3}\theta }}{/tex}
{tex} = \frac{{{{\sin }^2}\theta {{\cos }^2}\theta \left( {{{\sin }^3}\theta + {{\cos }^3}\theta } \right)}}{{\left( {{{\sin }^3}\theta + {{\cos }^3}\theta } \right)}}{/tex}
{tex} = {\sin ^2}\theta {\cos ^2}\theta {/tex}
LHS = RHS
Hence proved.