Question

Prove that 1+secA/secA =sin2a/1_cosA

Answer 1

L.H.S= (1+secA)/secA 

={1+(1/cosA)} / (1/cosA) 

= {(1+cosA)/cosA} / (1/cosA) 

={(1+cosA)/cosA} * (cosA/1) 

=1+cosA 

ON MULTIPLYING AND DIVIDING WITH 1-cosa

=(1+cosA)(1-cosA) / (1-cosA) 

={1- (cosA)^2} / (1-cosA) 

=(sinA)^2/(1-cosA). 

Answer 2

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