PROVE THAT : [sinA /1-cosA - 1-cosA/sinA ][cosA/1+sinA + 1+sinA/cosA] = 4cosecA.?

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Answer 1 · 2016-10-24T00:09:42

$LHS=[sinA/(1-cosA)-(1-cosA)/sinA][cosA/(1+sinA)+(1+sinA)/cosA]$

Now 1st part= $sinA/(1-cosA) -(1-cosA)/sinA$

$=(sinA(1+cosA))/((1-cosA)(1+cosA))-(1-cosA)/sinA$

$=(sinA(1+cosA))/(1-cos^2A) -(1/sinA-cosA/sinA)$

$=(sinA(1+cosA))/sin^2A-cscA+cotA$

$=(1+cosA)/sinA -cscA+cotA$

$=1/sinA+cosA/sinA -cscA+cotA$

$=cscA+cotA -cscA+cotA$

$=2cotA$

2nd part

$=cosA/(1+sinA)+(1+sinA)/cosA$

$=(cosA(1-sinA))/((1+sinA)(1-sinA))+(1/cosA+sinA/cosA)$

$=(cosA(1-sinA))/(1-sin^2A)+(secA+tanA)$

$=(cosA(1-sinA))/cos^2A+(secA+tanA)$

$=(1-sinA)/cosA+(secA+tanA)$

$=1/cosA-sinA/cosA+secA+tanA$

$=secA-tanA+secA+tanA$

$=2secA$

Whole LHS

$=2cotAxx2secA$

$=(4cosA)/sinAxx1/cosA$

$=4/sinA=4cscA=RHS$

Proved