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If $\sec \theta+\tan \theta=\sqrt{3}$ then the positive value of $\sin \theta$

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

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

squaring on both sides we get

$\sin^2\theta=$

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  • $\begingroup$ One should note that $$\sec\theta+\tan\theta= \tan\left( \frac\pi4 + \frac\theta2 \right).$$ $\endgroup$ – Michael Hardy Sep 16 '16 at 23:13
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Better: write it as $$\sin\theta+1=\sqrt3\,\cos\theta$$ and then square to obtain a quadratic in $\sin\theta$. Can you finish it from here?

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  • $\begingroup$ $sin^2\theta+1+2sin\theta=3cos^\theta$ $sin^2\theta+1+2sin\theta=3-3sin\theta$ $4sin^2\theta+2sin\theta-2$ $\endgroup$ – jyothika1 Sep 1 '14 at 5:41
  • $\begingroup$ Looks good, now solve the quadratic - if necessary, substitute $x=\sin\theta$. $\endgroup$ – David Sep 1 '14 at 5:45
  • $\begingroup$ -1,1/2 is it correct. $\endgroup$ – jyothika1 Sep 1 '14 at 5:48
  • $\begingroup$ $ 2x^2+x-1=0 $. $\endgroup$ – jyothika1 Sep 1 '14 at 5:49
  • $\begingroup$ That's correct - remember the question asked for the positive solution only, so $\sin\theta=\frac{1}{2}$. $\endgroup$ – David Sep 1 '14 at 6:27
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Since $1 = \sec^2 \theta - \tan^2 \theta = (\sec \theta + \tan \theta)(\sec \theta - \tan \theta)$, we have $\sec \theta - \tan \theta = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$.

Now, we have two equations:

(1) $\sec \theta + \tan \theta = \sqrt{3}$

(2) $\sec \theta - \tan \theta = \dfrac{\sqrt{3}}{3}$

Adding the two gives $\sec \theta = \dfrac{2\sqrt{3}}{3}$. Subtracting the 2nd from the 1st gives $\tan \theta = \dfrac{\sqrt{3}}{3}$.

Can you find $\sin \theta$ from this? Note that this method doesn't yield any extraneous solutions.

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  • $\begingroup$ $sec\theta/tan \theta$=1/2 $\endgroup$ – jyothika1 Sep 1 '14 at 5:59
  • $\begingroup$ Yes, that is correct. By the way, typing "\sec \theta" instead of "sec \theta" makes the $\LaTeX$ output look nicer. $\endgroup$ – JimmyK4542 Sep 1 '14 at 6:01
  • $\begingroup$ @JimmyK4542 Nice solution! But the question you meant to ask should be "Can you find sinθ from this?". $\sin\theta=\frac{\tan\theta}{\sec\theta} = 1/2$ $\endgroup$ – mike Sep 1 '14 at 6:18
  • $\begingroup$ @JimmyK4542, My favourite way (+1) $\endgroup$ – lab bhattacharjee Sep 2 '14 at 17:53
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Beginning with $1+\sin \theta = \sqrt{3} \cos \theta$, we can use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to obtain $$1+\sin \theta = \sqrt{3(1-\sin^2\theta)} \\ \Rightarrow 1 + 2\sin \theta + \sin^2 \theta = 3-3 \sin^2 \theta \\ \Rightarrow 4\sin^2 \theta + 2 \sin \theta - 2 = 0$$ Now you have an quadratic equation for $\sin \theta$ which you can solve by factoring so: $$4\sin^2 \theta + 2 \sin \theta - 2 = 0 \\ \Rightarrow 4(\sin \theta + 1)(\sin \theta - \frac{1}{2})=0 \\ \Rightarrow \sin \theta = -1, \sin \theta = \frac{1}{2}$$ Since we want the positive value, we conclude that $\sin \theta = 1/2$

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