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Trigonometry
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simplify 1+sinx/cosx+cosx/1+sinx
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\[\frac{1+\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}\] right?
take alcium [(1+sinx)(1+sinx)+(cosx.cosx)]/(cosx.(1+sinx))
[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
is that fine now @antonyo
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