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If tanΘ= -4/3 and Θ lies in the second quadrant, then sinΘ-2cosΘ=?

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sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???
why we need to divided by cosΘ??
you can't just divide by cos theta, you need to multiply also.

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Other answers:

i would suggest you find sec theta first, from the identity sec^2 theta = 1+ tan^2 theta and then, cos = 1/ sec, sin = tan * cos.
i have learnt sec yet, i have learnt sin cos tan only so far
there is a picture of an angle whose tangent is \(\frac{4}{3}\)
you need the third side, which you get via pythagoras or by remembering the 3 - 4 - 5 right triangle
now you see that \(\sin(\theta)=\frac{4}{5}\) and \(\cos(\theta)=\frac{3}{5}\) except that since you are in quadrant II you have \(\cos(\theta)=-\frac{3}{5}\) because in quadrant II cosine is negative
you want \[\sin(\theta)-2\cos(\theta)\] and you know the numbers that you need \[\frac{4}{5}-2\times (-\frac{3}{5})=\frac{4}{5}+\frac{6}{5}\]
rather easy if you draw a triangle otherwise a pita
Alternate way to look at the same thing, |dw:1365910821672:dw| so the angle inside triangle is \(\pi - \theta\) so, \(\cos(\pi-\theta)=3/5, -\cos \theta=3/5 , \cos \theta = -3/5\)
thank you both of you
and the next questions is : [1/cosθ + tanθ](1-sinθ)=?
write tan = sin/cos.
i got the answer -cosθ but there are only five choices: A. sinθ B. cosθ C. cos^(2)θ D. 1+sinθ E. sinθtanθ
how you got -cos theta ?
more specifically, how you got that - *minus*
\[\large \tan \theta=-\frac{4}{3}\] \[\large \theta =126`52'\] \[\sin \theta -2cos\theta=\sin 126`52' -2\cos126`52'\] \[\large =2\]
in this case \(a=\cos(\theta)\) and \(b=\sin(\theta)\) so you get \[\frac{1-\sin^2(\theta)}{\cos(\theta)}\] since \(1-\sin^2(\theta)=\cos^2(\theta)\) you end up with \[\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)\]
ummm, may be i got something wrong
the algebra is easier if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\)
then you won't get so confused when you multiply stuff out almost all the steps are algebra, there is very little trig here
\[(\frac{1}{\cos(\theta)}+\tan(\theta))(1-\sin(\theta))\] becomes the more simple to compute \[(\frac{1}{a}+\frac{b}{a})(1-b)\]
or if you prefer \[(\frac{1+b}{a})(1-b)\]
now you get \(\frac{1-b^2}{a}\) almost instantly
ohhh, i write sin^(2)θ-1 ....
actually \(1-\sin^2(\theta)\) for the numerator
i want to ask (1+sinθ)(1-sinθ)=1-sin^(2)θ??
ohhhh...then the answer is cosθ
yes, absolutely.
and the next question: If cosθ=1/k and 0°≤θ≤90°, then -tan(90°-θ)=?
|dw:1365911889934:dw| find the unknown side by using pythagoras theorem ?
|dw:1365911994361:dw| now find (tan 90-theta) = opposite side / adjacent side
sorry, that should read \(\large k^2-1\)
tan(90-theta)=-1/tan theta?
you can use that identity , its just another way to solve the same problem.
you'll get the same answer using any of the 2 methods.
\[-\frac{ 1 }{ \sqrt{k ^{2}}-1 }\] ??
if you meant , \(\Large -\frac{ 1 }{ \sqrt{k ^{2}-1} }\) then you are correct.
welcome ^_^

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