Prove that #(cosA-sinA+1)/(cosA+sinA-1)=cotA+cscA#?

2 Answers
Shwetank Mauria
Jan 27, 2018

Please see below.

Explanation:

#(cosA-sinA+1)/(cosA+sinA-1)#

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

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

= #((cosA+1)^2-sin^2A)/((cosA+sinA)^2-1)#

= #(cos^2A+2cosA+1-sin^2A)/(cos^2A+sin^2A+2sinAcosA-1)#

= #(2cos^2A+2cosA)/(2sinAcosA)#

= #cotA+cscA#

P dilip_k
Jan 28, 2018

#LHS=(cosA-sinA+1)/(cosA+sinA-1)#

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

#=(cosAsinA-sin^2A+sinA)/((cosA+sinA-1)sinA)#

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

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

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

#=(1+cosA)/sinA#

#=1/sinA+cosA/sinA#

#=cscA+cotA=RHS#

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