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Here are my steps

$$\dfrac{\sin2\theta+\cos2\theta+1}{\sin2\theta-\cos2\theta+1} = \cot\theta$$

$$\dfrac{2\sin\theta\cos\theta+2\cos^2\theta-1+1}{2\sin\theta\cos\theta-2\cos^2\theta-1+1} = \cot\theta$$

$$\dfrac{2\sin\theta\cos\theta+2\cos^2\theta}{2\sin\theta\cos\theta+2\cos^2\theta} =\cot\theta$$

Any help?

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  • $\begingroup$ Is there a mistake on second line- at denominator do the ones cancel? $\endgroup$
    – WindSoul
    Mar 12, 2023 at 17:03
  • $\begingroup$ @WindSoul yes $\cos2\theta$ is $2cos^2\theta-1$ $\endgroup$ Mar 12, 2023 at 17:04
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    $\begingroup$ Following on with WindSoul's comment: $-\cos 2\theta=-2\cos^2 \theta+1$ not $-2\cos^2\theta-1$ $\endgroup$ Mar 12, 2023 at 17:05
  • $\begingroup$ @Semiclassical my bad $\endgroup$ Mar 12, 2023 at 17:06
  • $\begingroup$ @Semiclassical I would appreciate a full solution for it if possible thanks $\endgroup$ Mar 12, 2023 at 17:07

1 Answer 1

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Using $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and $\cos(2\theta)=2\cos^2(\theta)-1=1-2\sin^2(\theta)$, we get

$$\require{cancel}\frac{\sin(2\theta)+\cos(2\theta)+1}{\sin(2\theta)-\cos(2\theta)+1}=\frac{2\sin(\theta)\cos(\theta)+2\cos^2(\theta)\cancel{-1+1}}{2\sin(\theta)\cos(\theta)-(\cancel{1}-2\sin^2(\theta))\cancel{+1}}$$ $$\require{cancel}=\frac{2\sin(\theta)\cos(\theta)+2\cos^2(\theta)}{2\sin(\theta)\cos(\theta)+2\sin^2(\theta)}=\frac{\cancel{2}\cos(\theta)\cdot \cancel{(\sin(\theta)+\cos(\theta))}}{\cancel{2}\sin(\theta)\cdot \cancel{(\sin(\theta)+\cos(\theta))}}=\cot(\theta)$$

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  • $\begingroup$ I don't understand the last step, can you clarify please? $\endgroup$ Mar 12, 2023 at 17:11
  • $\begingroup$ We are left with $\frac{\cos(\theta)}{\sin(\theta)}$, which is (for me, by definition) $\cot(\theta)$.. $\endgroup$
    – HappyDay
    Mar 12, 2023 at 17:13
  • $\begingroup$ What if $\sin(\theta)+\cos(\theta) = 0$ and so the cancellation in the last step is not valid? $\endgroup$ Mar 12, 2023 at 17:13
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    $\begingroup$ I assume we are trying to prove this only when both sides are defined, otherwise the claim is clearly false... $\endgroup$
    – HappyDay
    Mar 12, 2023 at 17:17

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