If = a b , show that x - b a +b = a^2-b^2 a^2+b^2 .

Question

If $\tan\text{x}=\frac{\text{a}}{\text{b}},$ show that $\frac{\text{x}\sin\text{x - b}\cos\text{x}}{\text{a}\sin\text{x}+\text{b}\cos\text{x}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}.$

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Answer

Given $=\tan\text{x}=\frac{\text{a}}{\text{b}}$ To show: $\frac{\text{a}\sin\text{x}-\text{b}\cos\text{x}}{\text{a}\sin\text{x}+\text{b}\cos\text{x}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$. Since, $\tan\text{x}=\frac{\text{a}}{\text{b}}$ $\Rightarrow\frac{\sin\text{x}}{\cos\text{x}}=\frac{\text{a}}{\text{b}}$ $\Rightarrow\text{b}\sin\text{x}=\text{a}\cos\text{x}=\lambda$ (Say) $\Rightarrow\sin\text{x}=\frac{\lambda}{\text{b}}$ and $\cos\text{x}=\frac{\lambda}{\text{a}}$ How $\frac{\text{a}\sin\text{x}-\text{b}\cos\text{x}}{\text{a}\sin\text{x}+\text{b}\cos\text{x}}=\frac{\frac{\text{a}.\lambda}{\text{b}}-\frac{\text{b}.\lambda}{\text{a}}}{\frac{\text{a}.\lambda}{\text{b}}+\frac{\text{b}.\lambda}{\text{a}}}$ $=\frac{\lambda\Big(\frac{\text{a}}{\text{b}}-\frac{\text{b}}{\text{a}}\Big)}{\lambda\Big(\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}\Big)}$ $=\frac{\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}}{\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}}$ $=\frac{\frac{\text{a}^2-\text{b}^2}{\text{ab}}}{\frac{\text{a}^2+\text{b}^2}{\text{ab}}}$ $=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$ $\text{Proved}$

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