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: $0$
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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