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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(x-3)(x+4). -3,4 multiply to -12, and add to get 1
Apply the negative exponent rule, so move the expression to the denominator to make the exponents positive.
Your answer is:
1
--------------
r^7 + s^12
I hope this helps you
6y=13-2x
y=13-2x/6
y=13/6-1/3x
The general form of the geometric sequence is
, where a sub n is the number term you're looking for (we're looking for the tenth term). a sub 1 is the first term in the sequence (ours is -6), r is the common ratio, and n-1 is the numbered term you're looking for minus 1. Our formula then looks like this:
. Simplify it to
. Take 2 to the 9th power then multiply it by -6 to get -3072. C is your answer.
