How do you prove that #(tanx+cosx)/(1+sinx)=secx#?

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How do you prove that #(tanx+cosx)/(1+sinx)=secx#?

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Answer 1 · 2015-02-14T13:32:15

I suspect that this identity doesn't work.
I tried to substitute an angle for #x# and doesn't work (I used #x=pi/6#, so you get #0.962=1.154#).
Maybe there is a typing error in it.

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How do you prove that #(tanx+cosx)/(1+sinx)=secx#?

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