Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved
We have to prove that
2/√3cosx + sinx = sec(Π/6-x)
To prove this we will solve the right-hand side of the equation which is
R.H.S = sec(Π/6-x)
= 1/cos(Π/6-x)
[As secƟ = 1/cosƟ)
= 1/[cos Π/6cosx + sin Π/6sinx]
[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]
= 1/[√3/2cosx + 1/2sinx]
= 1/(√3cosx + sinx]/2
= 2/√3cosx + sinx
R.H.S = L.H.S
Hence 2/√3cosx + sinx = sec(Π/6-x) is proved
Learn more about trigonometry here : brainly.com/question/7331447
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I’m not sure what the answer is but I hope someone can help you
I have no idea how to do this problem i dont know what a is
Answer:
5
Step-by-step explanation:
Hello There!
The mean is pretty much the average
So to find the mean we first add up all of the values
2+3+6+6+7+8+8=40
Then we count the amount of points
There are 8 points
our final step is to divide the total value by the total number of points
40/8=5
so the mean (average) would be 5