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Trigonometric Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions3 Marks
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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