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I am confused with the different methods of finding tirg solutions and always ending up getting a different and often wrong answer from the real one.

For example:

$cot\theta +cosec\theta =\sqrt{3}$

What I did is:

$cot\theta +cosec\theta =\sqrt{3}$

->$\frac{cos\theta }{sin\theta }+\frac{1}{sin\theta }=\sqrt{3}$

$cos\theta +1=\sqrt{3}sin\theta $

$cos\theta -\sqrt{3}sin\theta =-1$

Taking r=$\sqrt{\sqrt{3}^2+1}=2$ and $\sqrt{3}$=rsinx and 1=rcosx we have

$rcosxcos\theta -rsinxsin\theta =r\left(cos\left(x+\theta \right)\right)=2\left(cos\left(x+\theta \:\right)\right)$=$-1$

ie cos(x+$\theta $)=-1/2

which gives a solution as x+$\theta $=2n$\pi $+$\pi $

ie $\theta $=$2n\pi +\pi -\frac{\pi }{3}=2n\pi +\frac{2\pi }{3}$ (here x =$\frac{\pi }{3}$ as rsinx=2sinx=$\sqrt{3}$ ie sinx=$\sqrt{3}$/2 or x=$\frac{\pi }{3}$

which is completely different from the solution given in my text $\theta =2n\pi +\frac{\pi }{3}$ or is it? I dont understand

My text did the problem by taking $\frac{\left(1+sin\theta \right)}{cos\theta }=\frac{\left(cos\left(\frac{\theta \:}{2}\right)+sin\left(\frac{\theta \:}{2}\right)\right)^2}{cos^2\left(\frac{\theta \:\:}{2}\right)-sin^2\left(\frac{\theta \:\:}{2}\right)}=\frac{1+tan\left(\frac{\theta \:}{2}\right)}{1-tan\left(\frac{\theta \:}{2}\right)}=tan\left(\frac{\pi }{4}+\frac{\theta }{2}\right)\:and\:proceeding$

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  • $\begingroup$ Wrong, $\sqrt{\sqrt3+1}\neq2$. And $\cos(x+\theta)=-\frac12$ dos not imply $x+\theta=2n\pi+\pi$. $\endgroup$ – Yves Daoust Aug 7 '14 at 9:35
  • $\begingroup$ sorry i changed that to $\sqrt{\sqrt{3}^2+1}=2$ $\endgroup$ – Hijaz Aslam Aug 7 '14 at 9:46
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$$\cos(t)-\sqrt{3}\sin(t)=-1\longrightarrow \frac{1}2\cos(t)-\frac{\sqrt{3}}2\sin(t)=-\frac{1}2=\cos(\frac{2\pi}3)$$ Now to what trigonometric identity does LHS look like, when you know that $\cos(\pi/3)=1/2$ and $\sin(\pi/3)=\sqrt{3}/2$?

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HINT:

$$\csc\theta+\cot\theta=\sqrt3\iff\csc\theta-\cot\theta=\frac1{\sqrt3}$$

Add & subtract to find $\displaystyle\csc\theta=\frac2{\sqrt3}\iff\sin\theta=\frac{\sqrt3}2>0$ and $\displaystyle\cot\theta=\frac1{\sqrt3}\iff\tan\theta=\sqrt3>0$

So, $\theta$ must lie in the first Quadrant $\ \ \ \ (1)$

and $\displaystyle\tan\theta=\sqrt3=\tan\frac\pi3\implies\theta=n\pi+\frac\pi3$ where $n$ is an integer

$(1)\implies n$ must be even(why?)

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$\cos(x+\theta)=-1/2\Rightarrow x+\theta=2n\pi +\pi$ is not correct.

Since we have $\cos(\theta+(\pi/3))=-1/2,$ we have $$\theta +\frac{\pi}{3}=\frac{2}{3}\pi+2n\pi\Rightarrow \theta=\frac{\pi}{3}+2n\pi$$ or $$\theta +\frac{\pi}{3}=\frac{4}{3}\pi+2n\pi\Rightarrow \theta=\pi+2n\pi.$$ However, the latter is not valid (why?).

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  • $\begingroup$ @HijazAslam: Is there anything unclear? $\endgroup$ – mathlove Aug 7 '14 at 9:44
  • $\begingroup$ @HijazAslam: You mean you don't understand why $x=\pi/3$? Or you don't understand why $\cos(\theta+\pi/3)=-1/2\Rightarrow \theta+\pi/3=\theta=\pi/3+2n\pi\ \text{or}\ 4\pi/3+2n\pi$? $\endgroup$ – mathlove Aug 7 '14 at 9:52
  • $\begingroup$ Mathlove thanks I got my mistake but isn't the general solution of cosine gives $cos\left(x+\theta \right)=2n\pi \pm \frac{2\pi }{3}$ $\theta =2n\pi \pm \frac{2\pi }{3}-\frac{\pi }{3}$ $\theta =2n\pi +\frac{\pi }{3}$ and $\theta =2n\pi -\pi $ how is it $\theta =2n\pi +\pi $?? and why isn't it valid? $\endgroup$ – Hijaz Aslam Aug 7 '14 at 10:01
  • $\begingroup$ @HijazAslam: Sorry, I don't get what you mean. (By the way, the latter is not valid because if $\theta=\pi +2n\pi$, then neither $\cot\theta$ nor $\csc\theta$ can be defined.) $\endgroup$ – mathlove Aug 7 '14 at 10:06
  • $\begingroup$ @HijazAslam: $\cos (x+\theta)\not =2n\pi\pm (2\pi/3)$. note that $\cos(x+\theta)=-1/2.$ Do you understand why $x=\pi/3$? $\endgroup$ – mathlove Aug 7 '14 at 10:26

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