(tanA+cosecB)^2—(cotB–secA)^2=2tanAcotB(cosecA+secB)

We have,
LHS = (tanA + cosec B)² - (cotB - sec A)²
{tex}\Rightarrow{/tex} LHS = (tan²A + cosec²B + 2tanA cosecB ) - (cot²B + sec²A - 2cotB secA)
{tex}\Rightarrow{/tex} LHS = (tan²A - sec²A)+(cosec²B - cot²B)+2tanA cosecB +2cotB secA
But,  Sec²A - tan²A =1 & cosec²A - cot² A = 1
{tex}\therefore{/tex} LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA
{tex}\Rightarrow{/tex} LHS = 2 (tanA cosecB + cotB secA)
{tex}\Rightarrow{/tex} LHS = 2 tanA cotB{tex}\left( \frac { cosec\: B } { \cot B } + \frac { \sec A } { \tan A } \right){/tex} [Dividing and multiplying by tanA cotB]
{tex}\Rightarrow{/tex} LHS = 2tan A cotB{tex}\left\{ \frac { \frac { 1 } { \sin B } } { \frac { \cos B } { \sin B } } + \frac { \frac { 1 } { \cos A } } { \frac { \sin A } { \cos A } } \right\}{/tex} [Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]
{tex}\Rightarrow{/tex} LHS = 2 tanA cotB{tex}\left( \frac { 1 } { \cos B } + \frac { 1 } { \sin A } \right){/tex} = 2tanA cotB ( secB + cosecA ) = RHS.  Hence, proved.