prove the following trigonometric identities. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x) tan(x/2)=sin(x)/(1+cos(x))

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prove the following trigonometric identities. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x) tan(x/2)=sin(x)/(1+cos(x))

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\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x) \tan(x/2)=\frac{ \sin (x) }{1+\cos (x) } \] is that your question?
no they are two separate equations. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x) tan(x/2)=sin(x)/(1+cos(x))

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Other answers:

oK
I will do only one, which one do you like?
do the top one
you can do either one if you want
for the second \[tan\frac{x}{2}=\frac{sinx}{1+cosx}\] try this \[tan\frac{x}{2}=\frac{2sin\frac{x}{2}cos \frac{x}{2}}{1+(1-2sin^2 \frac{x}{2})}\]
sec (pi/6 - x) = 1 / (sqrt3/2 cos x + sin pi/6 sin x) = 2 / (sqrt 3 cos + sin x)
* I missed out the second step which is = 1 / (cos pi/6 cos x + sin pi/6 sin x)
\[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x)\] \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6-x)} }\] using:cos(u-v)=cos u cos v+sin u sin v \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6) \times \cos(x)}+\sin{(\pi/6) \times \sin(x)} }\]
i got that far ASAAD123 put i didn't know what to do after that
the denominator = cos (pi/6 - x)
and 1 / cos = sec
wouldn't you also change sin(pi/6) to 1/2
no its the sin of a compound angle (pi/6 - x) not sin pi/6
ok
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so how do you get the sides to equal
i had to use sin pi/6 = 1/2 because i started with RHS and converted to LHS whereas ASAAD did the reverse
with these identities you choose one side and try to convert it to the other. This proves the identity.
ok
i think i understand it now
so you can use asaad's or mine. Either would do.
There is only more more step to prove IrishBoy's solution
i already got irish boy's solution

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