Question

CosecA+secA=cosecB+secBProves that tana+b/2=cotA cotB

Answer 1

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

$\rm :\longmapsto\:cosecA + secA = cosecB + secB$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:tan\bigg(\dfrac{A + B}{2} \bigg) = cotA \: cotB$

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

$\underbrace{\boxed{ \tt{ \:cosecx = \frac{1}{sinx} }}}$

$\underbrace{\boxed{ \tt{ \:secx = \frac{1}{cosx} }}}$

$\underbrace{\boxed{ \tt{ cotx\: = \frac{cosx}{sinx} }}}$

$\underbrace{\boxed{ \tt{ tanx\: = \frac{sinx}{cosx} }}}$

$\underbrace{\boxed{ \tt{cosx - cosy = 2sin\bigg(\dfrac{y + x}{2} \bigg)sin\bigg(\dfrac{y - x}{2} \bigg) \: }}}$

$\underbrace{\boxed{ \tt{sinx - siny = 2cos\bigg(\dfrac{y + x}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg) \: }}}$

$\purple{\large\underline{\bf{Solution-}}}$

Given that

$\rm :\longmapsto\:cosecA + secA = cosecB + secB$

On transposition, we get

$\rm :\longmapsto\:cosecA - cosecB = secB - secA$

$\rm :\longmapsto\:\dfrac{1}{sinA} - \dfrac{1}{sinB} = \dfrac{1}{cosB} - \dfrac{1}{cosA}$

$\rm :\longmapsto\:\dfrac{sinB - sinA}{sinAsinB} = \dfrac{cosA - cosB}{cosAcosB}$

On applying alternendo, we get

$\rm :\longmapsto\:\dfrac{cosAcosB}{sinAsinB} = \dfrac{cosA - cosB}{sinB - sinA}$

$\rm :\longmapsto\:cotA \: cotB = \dfrac{2sin\bigg(\dfrac{A + B}{2} \bigg)sin\bigg(\dfrac{B - A}{2} \bigg)}{2cos\bigg(\dfrac{A + B}{2} \bigg)sin\bigg(\dfrac{B - A}{2} \bigg)}$

$\rm :\longmapsto\:cotA \: cotB = tan \bigg(\dfrac{A + B}{2} \bigg)$

Hence, Proved

Additional Information :-

$\underbrace{\boxed{ \tt{cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \: }}}$

$\underbrace{\boxed{ \tt{sinx + siny = 2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \: }}}$

$\underbrace{\boxed{ \tt{2sinxcosy = \: sin(x + y) + sin(x - y)}}}$

$\underbrace{\boxed{ \tt{2cosxcosy = \: cos(x + y) + cos(x - y)}}}$

$\underbrace{\boxed{ \tt{ 2sinxsiny\: = \: cos(x - y) - cos(x + y)}}}$