#I need help with these questions

21 messages · Page 1 of 1 (latest)

supple hollow
prime dew
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1st one you get $\frac{\cos(t)(\sin(t)-1)+\cos(t)(\sin(t)+1)}{(\cos(t)-1)(\cos(t)+1)}$ which you then expand and simplify

obtuse cipherBOT
prime dew
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second one you Binomially expand:
1/cosy = secy
sec²y+2secytany+tan²y
2secytany = 2/cosy × siny/cosy
2siny/cos²y
all have denominators of cos²y now
(1+2siny+sin²y)/cos²y

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which is just (1+siny)²/cos²y
to which we rationalise our denominator on the RHS
(1+siny)(1+siny)/(1+siny)(1-siny)

(1+siny)²/(1-sin²y) = (1+siny)²/(cos²y)

chilly niche
chilly niche
prime dew
chilly niche
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forgot to mention it

prime dew
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no

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the question is correct

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sin²t-1≠cos²t

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other way around, that's where you dropped the minus

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sin²t-1=-cos²t

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@chilly niche

chilly niche
prime dew
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dw

supple hollow
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wait i’m confused

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i’m kinda lost at what i’m looking at i’m so sorry

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You’re supposed to fully expand it and then simply the solution?