## Trigonometry (11th Edition) Clone

$$\tan(\theta-45^\circ)+\tan(\theta+45^\circ)=2\tan2\theta$$ As 2 sides are equal, the equation has been verified to be an identity. The proof is below.
$$\tan(\theta-45^\circ)+\tan(\theta+45^\circ)=2\tan2\theta$$ We would look at the left side first. $$X=\tan(\theta-45^\circ)+\tan(\theta+45^\circ)$$ - Apply the tangent difference and sum identities to $X$, which state $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$ $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ $X$ would be $$X=\frac{\tan\theta-\tan45^\circ}{1+\tan\theta\tan45^\circ}+\frac{\tan\theta+\tan45^\circ}{1-\tan\theta\tan45^\circ}$$ $$X=\frac{\tan\theta-1}{1+\tan\theta\times1}+\frac{\tan\theta+1}{1-\tan\theta\times1}$$ $$X=\frac{\tan\theta-1}{1+\tan\theta}+\frac{\tan\theta+1}{1-\tan\theta}$$ $$X=\frac{(\tan\theta-1)(1-\tan\theta)+(1+\tan\theta)^2}{(1+\tan\theta)(1-\tan\theta)}$$ On the numerator, we expand everything. On the denominator, recall that $(A+B)(A-B)=A^2-B^2$ for $A=1$ and $B=\tan\theta$ here. $$X=\frac{(\tan\theta-\tan^2\theta-1+\tan\theta)+(1+\tan^2\theta+2\tan\theta)}{1-\tan^2\theta}$$ $$X=\frac{(\tan\theta+\tan\theta+2\tan\theta)+(-\tan^2\theta+\tan^2\theta)+(-1+1)}{1-\tan^2\theta}$$ $$X=\frac{4\tan\theta}{1-\tan^2\theta}$$ $$X=2\times\frac{2\tan\theta}{1-\tan^2\theta}$$ - $\frac{2\tan\theta}{1-\tan^2\theta}$ can be written into $\tan2\theta$, as stated in the Double-Angle Identity for tangent. $$X=2\tan2\theta$$ Therefore, $$\tan(\theta-45^\circ)+\tan(\theta+45^\circ)=2\tan2\theta$$ As 2 sides are equal, the equation has been verified to be an identity.