# How to prove $\frac{\sin2\theta+\cos2\theta+1}{\sin2\theta-\cos2\theta+1} = \cot\theta$

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?

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

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
$$\require{cancel}\frac{\sin(2\theta)+\cos(2\theta)+1}{\sin(2\theta)-\cos(2\theta)+1}=\frac{2\sin(\theta)\cos(\theta)+2\cos^2(\theta)\cancel{-1+1}}{2\sin(\theta)\cos(\theta)-(\cancel{1}-2\sin^2(\theta))\cancel{+1}}$$ $$\require{cancel}=\frac{2\sin(\theta)\cos(\theta)+2\cos^2(\theta)}{2\sin(\theta)\cos(\theta)+2\sin^2(\theta)}=\frac{\cancel{2}\cos(\theta)\cdot \cancel{(\sin(\theta)+\cos(\theta))}}{\cancel{2}\sin(\theta)\cdot \cancel{(\sin(\theta)+\cos(\theta))}}=\cot(\theta)$$
• We are left with $\frac{\cos(\theta)}{\sin(\theta)}$, which is (for me, by definition) $\cot(\theta)$.. Mar 12, 2023 at 17:13
• What if $\sin(\theta)+\cos(\theta) = 0$ and so the cancellation in the last step is not valid? Mar 12, 2023 at 17:13