Question : If $a \cot \theta=b$, what will be the value of $\frac{b \cos \theta–a \sin \theta}{b \cos \theta+a \sin \theta}$?
Option 1: $\frac{b^2+a^2}{b^2–a^2}$
Option 2: $b^2+a^2$
Option 3: $\frac{b^2–a^2}{b^2+a^2}$
Option 4: $0$
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Correct Answer: $\frac{b^2–a^2}{b^2+a^2}$
Solution : Given: $a \cot \theta=b$ $⇒\cot \theta=\frac{b}{a}$ $⇒\frac{\cos \theta}{\sin \theta}=\frac{b}{a}$ Let $\cos a = bk $ and $\sin a =ak$ (where $k$ is a constant) Putting these values in the expression, we get, $\frac{b \cos \theta–a \sin \theta}{b \cos \theta+a \sin \theta}$ $=\frac{b×bk-a×ak}{b×bk+a×ak}$ $=\frac{k(b^2–a^2)}{k(b^2+a^2)}$ $=\frac{b^2–a^2}{b^2+a^2}$ Hence, the correct answer is $\frac{b^2–a^2}{b^2+a^2}$.
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