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Trigonometry 13 Online
OpenStudy (anonymous):

simplify 1+sinx/cosx+cosx/1+sinx

hero (hero):

\[\frac{1+\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}\] right?

OpenStudy (anonymous):

take alcium [(1+sinx)(1+sinx)+(cosx.cosx)]/(cosx.(1+sinx))

OpenStudy (anonymous):

[1+sin(x)^2+2sinx+cosx^2]/cosx(1+sinx) [1+2sinx+(sinx^2+cosx^2)]/cosx(1+sinx).. as (sinx^2+cosx^2)=1 [1+2inx+1]/cosx(1+sinx) (2+2sinx)/cosx(1+sinx) 2(1+sinx)/cosx(1+sinx) 2/cosx 2.secx answer

OpenStudy (anonymous):

is that fine now @antonyo

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