trigonometry
I am trying to prove the identity below to help with the simplification of another function that I'm investigating as it doesn't appear to be a standard trig identity.
$$
\tan\left(x\right) + \tan\left( y \right) = \frac{{\sin\left( {x + y} \right)}}{{\cos\left( x \right)\cos\left( y \right)}}
$$
Any assistance gratefully appreciated.
For fun, I found a "trigonograph" of this identity (for acute $\theta$).
In the diagram, $\overline{AB}$ is tangent to the unit circle at $P$. The "trig lengths" (except for $|\overline{AQ}|$) should be clear.
We note that $\angle BPR \cong \angle RPP^\prime$, since these inscribed angles subtend congruent arcs $\stackrel{\frown}{PR}$ and $\stackrel{\frown}{RP^\prime}$. Very little angle chasing gives that $\triangle APQ$ is isosceles, with $\overline{AP} \cong \overline{AQ}$ (justifying that last trig length). Then, $$\triangle SPR \sim \triangle OQR \implies \frac{|\overline{SP}|}{|\overline{SR}|} = \frac{|\overline{OQ}|}{|\overline{OR}|} \implies \frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta+\tan\theta}{1}$$
Using some trig-identities we have: $$\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$$ and $$\cos(2x)=2\cos^2(x)-1$$ and $$\tan^2(x)=\sec^2(x)-1$$ we have (on the left hand side): $$\begin{align}\tan(2x)-\tan(x)&=\frac{2\tan(x)}{1-\tan^2(x)}-\tan(x)\\&=\frac{2\tan(x)-\tan(x)+\tan^3(x)}{1-\tan^2(x)}\\&=\tan(x)\frac{1+\tan^2(x)}{1-\tan^2(x)}\\&=\tan(x)\frac{1+\sec^2(x)-1}{1-\sec^2(x)+1}\\&=\tan(x)\left[\frac{2-\sec^2(x)}{\sec^2(x)}\right]^{-1}\\&=\tan(x)\left[2\cos^2(x)-1\right]^{-1}\\&=\tan(x)[\cos(2x)]^{-1}\\&=\tan(x)\sec(2x)\end{align}$$ and we are done.
Best Answer
You are asking about a proof of the identity $$ \tan\left(x\right) + \tan\left( y \right) = \frac{{\sin\left( {x + y} \right)}}{{\cos\left( x \right)\cos\left( y \right)}} $$
Using $\tan(x)=\frac{\sin(x)}{\cos(x)}$, we get $$\tan\left(x\right) + \tan\left( y \right) = \frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(y)}\\ =\frac{\sin(x)\cdot \cos(y) + \sin(y)\cdot \cos(x)}{\cos(x)\cdot \cos(y)}$$
Using the identity $\sin(x+y)=\sin(x)\cdot \cos(y) + \sin(y)\cdot \cos(x)$ gives you the answer of your question.