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Simplify the expression. cos x + sin x tan x
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is it sec x ?
Yes, yes it is!
Thank you Batmaan!
\[cosx+sinxtanx = cosx+sinx (\frac{ sinx }{ cosx })\]
\[\large \rm \dfrac{adj}{hyp} + \dfrac{opp}{hyp} \dfrac{opp}{adj}\]
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\[cosx+\frac{ \sin^2x }{ cosx } \implies \frac{ \cos^2x+\sin^2x }{ cosx } \implies secx\]
I now understand. Thank you both for the help/ explanation :)
Yeah, just remember the identity cos^2x+sin^2x, so 1/cosx = secx
cos^2x+sin^2x = 1*
Also, welcome back :P
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I will! and awesome to see you
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