Question

sin/1-cos+tan/1+cos=sec cosec+cot​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\bf \: Prove \: that : \sf \: \dfrac{sinx}{1 - cosx} + \dfrac{tanx}{1 + cosx} = secx \: cosecx \: + cotx$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$1. \: \: \: \boxed{ \bf{tanx = \dfrac{sinx}{cosx}}}$

$2. \: \: \: \boxed{ \bf{ {sin}^{2} x + {cos}^{2} x = 1}}$

$3. \: \: \: \boxed{ \bf{secx \: = \: \dfrac{1}{cosx} }}$

$4. \: \: \: \boxed{ \bf{cosecx \: = \: \dfrac{1}{sinx} }}$

$5. \: \: \: \boxed{ \bf{cotx \: = \: \dfrac{cosx}{sinx} }}$

$\large\underline{\sf{Solution-}}$

Consider LHS,

$\sf \: \dfrac{sinx}{1 - cosx} + \dfrac{tanx}{1 + cosx}$

$= \: \sf \: \dfrac{sinx}{1 - cosx} + \dfrac{sinx}{cosx(1 + cosx)}$

$= \: \sf \: sinx \: \bigg(\dfrac{1}{1 - cosx} + \dfrac{1}{cosx \: (1 + cosx)} \bigg)$

$= \: \sf \: sinx \: \bigg(\dfrac{cosx(1 + cosx) + 1 - cosx}{cosx(1 + cosx)(1 - cosx)} \bigg)$

$= \: \sf \: sinx \: \bigg(\dfrac{ \cancel{cosx} + {cos}^{2} x + 1 - \cancel{cosx}}{cosx(1 - {cos}^{2}x )} \bigg)$

$= \: \sf \: sinx\bigg( \dfrac{1 + {cos}^{2}x }{cosx \: {sin}^{2}x } \bigg)$

$= \: \sf \: \dfrac{1 + {cos}^{2}x }{sinx \: cosx}$

$= \: \sf \: \dfrac{1}{sinx \: cosx} + \dfrac{ {cos}^{2}x }{sinx \: cosx}$

$= \: \sf \: secx \: cosecx \: + \: \dfrac{cosx}{sinx}$

$= \: \sf \: secx \: cosecx \: + \: cotx$

$\bf \: = \: RHS$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \bf{ {sec}^{2} x - {tan}^{2} x = 1}}$

$\boxed{ \bf{ {cosec}^{2} x - {cot}^{2} x = 1}}$

$\boxed{ \bf{sin(90\degree \: - x) = cosx}}$

$\boxed{ \bf{cos(90\degree \: - x) = sinx}}$

$\boxed{ \bf{tan(90\degree \: - x) = cotx}}$

$\boxed{ \bf{cot(90\degree \: - x) = tanx}}$

$\boxed{ \bf{sec(90\degree \: - x) = cosecx}}$

$\boxed{ \bf{cosec(90\degree \: - x) = secx}}$