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Question Number 187491 by cortano12 last updated on 18/Feb/23

$$\{\begin{array}{l}\sin x+\sin y=a\\ \cos x+\cos y=b\end{array}$$$$\tan x+\tan y=?$$

Answered by horsebrand11 last updated on 18/Feb/23

$$(\ast )\tan \left(\frac{x-y}{2}\right)=\frac{a}{b}$$$$\Rightarrow \tan (x+y)=\frac{2ab}{b^2-a^2}$$$$\Rightarrow \cos (x+y)=\frac{b^2-a^2}{b^2+a^2}$$$$\Rightarrow \cos (x-y)=\frac{a^2+b^2-2}{2}$$$$\Rightarrow \cos x\cos y=\frac{{(a^2+b^2)}^2-4a^2}{4(a^2+b^2)}$$$$(\therefore )\tan x+\tan y=\tan (x+y)[1-\tan x\tan y]$$$$=\tan (x+y)\left[\frac{\cos (x+y)}{\cos x\cos y}\right]$$$$=\left(\frac{2ab}{b^2-a^2}\right)\left(\frac{b^2-a^2}{b^2+a^2}\right)\left(\frac{4(a^2+b^2)}{{(a^2+b^2)}^2-4a^2}\right)$$$$=\frac{8ab}{{(a^2+b^2)}^2-4a^2}$$

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