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
#SPJ9
Answer: 41.832
Step-by-step explanation:
I like to write decimals without decimals when multiplying. for this I would do
3486 x 12 which equals 41832 then You would move the decimal up from 3 spaces from the 2 from 41832 which would put the decimal between the 1 and 8
what's up? the answer to this is 339+340+341
best of luck with your studies
Your answer is highlighted!
~Good Luck~
Answer:
8 mins
Step-by-step explanation:
First you see how many mins he can read on 1 page than you divided the number by 24 to get your answer.
