How do you solve ( tanx+secx)^2 = (1+sinx)/(1-sinx)(tanx+secx)2=1+sinx1−sinx?

1 Answer
Nghi N.
May 23, 2015

(sin x/cos x - 1/cos x) ^2 = (sin x - 1)^2/(cos^2 x) =(sinxcosx−1cosx)2=(sinx−1)2cos2x=

= (sin x - 1)^2/(1 - sin^2x) =[ (sin x - 1)(sin x - 1)]/[(sin x-1)(sinx + 1)] (sinx−1)21−sin2x=(sinx−1)(sinx−1)(sinx−1)(sinx+1)=

= [-(1 + sin x)]/[-(1 - sin x)]−(1+sinx)−(1−sinx) = (1 + sin x)/(1 - sin x)1+sinx1−sinx

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