Answered respectively;
1) 0.13514
2) 0.125
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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<span>8(3x + 6) + 13---expand using distributive property
=24x + 48+13 ...combine like terms
=24x +61....simplify
answer is C.24x + 61
hope it helps</span>
I don’t know, I forgot how to solve problems like this, look it up
Answer:
it is B.
Step-by-step explanation:
B. can i have brainlyest please
