Question : If $\cos\theta = \frac{x^2\:-\:y^2}{x^2\:+\:y^2}$, then the value of $\cot\theta$ is equal to:
Option 1: $\frac{2xy}{x^{2}\:-\:y^{2}}$
Option 2:
$\frac{2xy}{x^2\:+\:y^2}$
Option 3:
$\frac{x^{2}\:+\:y^{2}}{2xy}$
Option 4:
$\frac{x^{2}\:-\:y^{2}}{2xy}$
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Correct Answer:
$\frac{x^{2}\:-\:y^{2}}{2xy}$
Solution :
Let right angle triangle at B and $\angle C = \theta$.
$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{CB}{AC} = \frac{x^{2}\:-\:y^{2}}{x^{2}\:+\:y^{2}}$
Applying Pythagoras' theorem, $(x^{2}+y^{2})^{2} =(x^{2}-y^{2})^{2}+(\text{Height})^{2}$
$⇒x^{4}+y^{4}+2x^{2}y^{2} =x^{2}+y^{2}-2x^{2}y^{2} +(\text{Height})^{2}$
$⇒4x^{2}y^{2} = (\text{Height})^{2}$
$⇒\sqrt{4x^{2}y^{2}} = \text{Height}$
$⇒2xy = \text{Height}$
So, $\cot\theta = \frac{\text{Base}}{\text{Height}} = \frac{x^{2}\:-\:y^{2}}{2xy}$
Hence, the correct answer is $\frac{x^{2}\:–\:y^{2}}{2xy}$.
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