Question

show that
(secα+tanα+1/secα-tanα+1)^2=cosecα+1/cosec-1

Answer 1

LHS=(seca+tana+sec^2a-tan^2a/seca-tana+1)^2
=(seca+tana(1+seca-tana) /seca-tana+1)^2
=(seca+tana)^2
=(sec^2a+tan^2a+2secatana)
=(1/cos^2a +sin^2a/cos^2a +2sina/cos^2a)
=(1+sin^2a+2sina/cos^2a)
=(1+sina)^2/1-sin^2a
=(1+sina)(1+sina)/(1+sina)(1-sina)
=(1+sina)/(1-sina)
=(1+1/coseca)(1-1/coseca)
=(coseca+1)(coseca-1)=RHS

Answer 2

Answer:

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