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Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $

My attempt is as follows:-

$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$ $$\left(\sin\theta-\cos\theta\right)^2+\sin\theta\left(\sin\theta-\cos\theta\right)=0$$ $$\left(\sin\theta-\cos\theta\right)(2\sin\theta-\cos\theta)=0$$ $$\tan\theta=1 \text { or } \tan\theta=\frac{1}{2}$$ $$\theta=n\pi+\frac{\pi}{4}$$

For $\tan\theta=\dfrac{1}{2} \text { how to make use of given condition } \tan71 ^{\circ}34^{\prime}=3$

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Use $3=\frac{1+\frac12}{1-1\cdot\frac12}=\tan(45^\circ+\theta)$.

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  • $\begingroup$ but how did you do that, guessing or something else $\endgroup$ – user3290550 Oct 26 '19 at 16:53
  • $\begingroup$ @user3290550 I solved $3=\frac{k+\frac12}{1-\frac12 k}$ so I could use this. $\endgroup$ – J.G. Oct 26 '19 at 16:55
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As $1+2+3=1\cdot2\cdot3$

$\arctan1+\arctan2+\arctan 3=n\pi$

using Show $\tan(x)+\tan(y)+\tan(z) = \tan(x) \tan(y) \tan(z)$

As $\arctan(u)<\dfrac\pi2$

$n=1$

Now $\arctan1=\dfrac\pi4$

$\arctan2+\arctan\dfrac12=\arctan2+$arccot$2=\dfrac\pi2$

See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

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