How do you prove # (sec^2x-1) / (secx - 1) = secx + sin^2x+cos^2x #?

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Answer 1 · 2018-05-04T04:08:05

As proved

#R H S = sec x + sin^2x + cos^2 x#

#=> sec x + 1, " as "color(crimson)(sin^2x + cos^2x = 1#

#=> ((sec x + 1 ) * (sec x -1)) / (sec x - 1), color(crimson)(" multiply and divide by " (sec x - 1)#

#=> (sec^2x - 1) / (sec x - 1) = L H S#

Answer 2 · 2018-05-04T04:11:29

As proved

#(sec^2x - 1) / (sec x - 1)#

#((sec x + 1) * cancel(sec x - 1) ) / cancel(sec x - 1)#

#=> sec x + 1 = sec x + sin^2 x + cos^2 x " as " color(crimson)(sin^2 x + cos^2 x = 1#

#=> R H S#