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