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Core 3- Trig

[sec(x) - 1 ]/tan(x) = [ tan (x)]/ sec(x) +1

So, i have to prove this identity is true. Ive managed to show it by dividing both sides by one of the sides, however i cant think how to start from either the left or right, and work myself to the other side. Any help would be awesome :smile:
Reply 1
Avatar for TenOfThem
TenOfThem
18
sec−1tan=[1cos−1]cossin=1−cossin=1−cos2sin(1+cos)=sin1+cos=tansec+1\frac{sec-1}{tan} = [\frac{1}{cos} - 1]\frac{cos}{sin} = \frac{1-cos}{sin} = \frac{1-cos^2}{sin(1+cos)} = \frac{sin}{1+cos} = \frac{tan}{sec+1}
Reply 2
Avatar for raheem94
raheem94
13
Original post by antonio108
[sec(x) - 1 ]/tan(x) = [ tan (x)]/ sec(x) +1

So, i have to prove this identity is true. Ive managed to show it by dividing both sides by one of the sides, however i cant think how to start from either the left or right, and work myself to the other side. Any help would be awesome :smile:


secx−1tanx≡tanxsecx+1 \displaystyle \frac{secx-1}{tanx} \equiv \frac{tanx}{secx+1}

LHS =
Unparseable latex formula:

\displaystyle \frac{secx-1}{tanx} = \frac{\frac1{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=\frac{1-cosx}{sinx}=\frac{(1-cosx)(1+cosx)}{sinx(1+cosx)}}


sin2xsinx(1+cosx)=sinx1+cosx=sinxcosx1+cosxcosx=tanxsecx+1 \displaystyle \frac{sin^2x}{sinx(1+cosx)} = \frac{sinx}{1+cosx}=\frac{\frac{sinx}{cosx}}{\frac{1+cosx}{cosx}}=\frac{tanx}{secx+1}
Reply 3
Avatar for steve2005
steve2005
17
Several methods, here is mine.




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Reply 4
Avatar for antonio108
antonio108
OP
1
Ah thank you very much, just looking through and slowly realising what i have to look out for.
Again, thanks!

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